NCERT solutions for Class 10 Maths Chapter 4 “Quadratic Equations” on the official website of NCERT or on other educational websites. You can also find various textbooks and online resources that offer solutions and explanations for the chapter. Here are some of the topics covered in class 10 math quadratic equation solution:
NCERT Solutions For Class 10 Maths Chapter 4 Summary
- Introduction to Quadratic Equations
- Standard Form of a Quadratic Equation
- Solutions of a Quadratic Equation
- Nature of Roots
- Quadratic Equation in Two Variables
You can use these resources to practice solving problems related to the chapter and improve your understanding of quadratic equations.
ax^2 + bx + c = 0
where a, b, and c are constants, and x is the variable.
The general solution to a quadratic equation is given by the quadratic formula:
x = (-b ± √(b^2 – 4ac)) / 2a
This formula gives the values of x that make the quadratic equation true. If the discriminant, b^2 – 4ac, is negative, then the quadratic equation has no real solutions, since the square root of a negative number is not a real number.
There are several methods for solving quadratic equations, including factoring, completing the square, and using the quadratic formula. Factoring involves finding two factors of the quadratic expression that multiply to give the quadratic expression itself. Completing the square involves manipulating the quadratic expression into a perfect square trinomial. Both methods can be used to solve quadratic equations with real solutions.
Quadratic equations have many applications in science, engineering, and mathematics. They are used, for example, to model the trajectory of projectiles, the shape of parabolic mirrors, and the behavior of vibrating systems.
What is Quadratic Equations
A quadratic equation is a second-degree polynomial equation in one variable of the form:
ax^2 + bx + c = 0
where x is the variable, and a, b, and c are constants, with “a” not equal to zero.
Quadratic equations have the degree of 2 because the variable is raised to the power of 2. This means that the graph of a quadratic equation is a parabola, which can either open upwards or downwards.
The quadratic equation can be solved using various methods such as factoring, completing the square, or by using the quadratic formula. The quadratic formula is given by:
x = (-b ± √(b^2 – 4ac)) / 2a
where ± means that there are two possible solutions, and the formula can be used to find the roots of a quadratic equation, which are the values of x that make the equation true.
Quadratic equations have many real-world applications, such as in physics, engineering, and finance. They can be used to model the motion of projectiles, the path of a satellite, or the cost and revenue functions of a business.
Use of Quadratic Equations
Quadratic equations have a wide range of practical applications in various fields, including science, engineering, finance, and economics. Here are some examples of how quadratic equations are used:
- Physics: In physics, quadratic equations are used to model the motion of projectiles, such as missiles or bullets. The equation helps calculate the trajectory of the projectile and predict its landing point.
- Engineering: Engineers use quadratic equations to design structures such as bridges and arches. Quadratic equations can help find the optimal shape for a structure to ensure its stability.
- Finance: Quadratic equations are used in finance to calculate the return on investment, the breakeven point, and other financial metrics. For example, the quadratic equation can be used to find the price at which a business should sell a product to maximize profit.
- Computer Graphics: Quadratic equations are also used in computer graphics to create 3D animations and special effects. By using quadratic equations, animators can create realistic animations of objects moving in space.
- Ecology: Quadratic equations can be used in ecology to model population growth. The equation can help predict the number of individuals in a population over time, taking into account factors such as birth rate and mortality rate.
- Statistics: Quadratic equations can be used in statistics to create regression models that help predict future trends based on past data.
In summary, quadratic equations have a wide range of applications in many fields, and their ability to model a wide range of phenomena makes them a valuable tool for scientists, engineers, and researchers.
Type of Quadratic Equations
There are several types of quadratic equations, which can be classified based on the properties of their coefficients or variables. Here are some of the most common types of quadratic equations:
- Standard Quadratic Equation: A standard quadratic equation is of the form ax^2 + bx + c = 0, where a, b, and c are constants, and x is the variable. This is the most general form of a quadratic equation.
- Quadratic Equation in Vertex Form: A quadratic equation in vertex form is of the form y = a(x-h)^2 + k, where a, h, and k are constants, and x and y are the variables. This form of the quadratic equation represents a parabola with the vertex at the point (h,k).
- Quadratic Equation in Intercept Form: A quadratic equation in intercept form is of the form y = a(x-x1)(x-x2), where a, x1, and x2 are constants, and x and y are the variables. This form of the quadratic equation represents a parabola with x-intercepts at x1 and x2.
- Quadratic Equation with Rational Coefficients: A quadratic equation with rational coefficients is an equation in which all the coefficients a, b, and c are rational numbers.
- Quadratic Equation with Irrational Coefficients: A quadratic equation with irrational coefficients is an equation in which one or more of the coefficients a, b, and c is an irrational number.
These are some of the types of quadratic equations that are commonly encountered in mathematics. Understanding the properties and characteristics of each type of equation can help in solving quadratic equations more effectively.
Quadratic Equations Solution
To solve a quadratic equation, we need to find the values of the variable that make the equation true. There are several methods to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. Here’s how each method works:
- Factoring: If the quadratic equation can be factored into two linear factors, then we can use the zero-product property to find the solutions. For example, the equation x^2 + 5x + 6 = 0 can be factored as (x + 2)(x + 3) = 0, which gives us the solutions x = -2 and x = -3.
- Completing the Square: If the quadratic equation cannot be factored, we can use the technique of completing the square to solve for the variable. To do this, we first rewrite the equation in the form a(x – h)^2 + k = 0, where h and k are constants. We can then solve for x by taking the square root of both sides of the equation. For example, the equation x^2 + 4x – 5 = 0 can be rewritten as (x + 2)^2 – 9 = 0, which gives us the solutions x = 1 and x = -5.
- Quadratic Formula: The quadratic formula is a general formula that can be used to find the solutions of any quadratic equation of the form ax^2 + bx + c = 0. The formula is x = (-b ± sqrt(b^2 – 4ac)) / 2a, where ± means that there are two possible solutions. For example, the equation 2x^2 + 3x – 5 = 0 can be solved using the quadratic formula, which gives us the solutions x = 1 and x = -2.5.
These are the main methods for solving quadratic equations. It is important to practice using each method to become proficient in solving quadratic equations of different types and complexity.
Quadratic Equations Ex 4.1 Examples
- x^2 – 6x + 5 = 0 This quadratic equation can be factored as (x-1)(x-5) = 0, so the solutions are x=1 and x=5.
- 2x^2 + 3x – 4 = 0 This quadratic equation can be solved using the quadratic formula: x = (-3 ± sqrt(3^2 – 4(2)(-4))) / (2(2)) = (-3 ± sqrt(25)) / 4. The solutions are x = -4/2 or x = 1/2.
- 5x^2 + 2x + 1 = 0 This quadratic equation has no real solutions because the discriminant is negative (b^2 – 4ac = 2^2 – 4(5)(1) = -16).
- (x+3)^2 – 4 = 0 This quadratic equation can be rewritten as (x+3)^2 = 4. Taking the square root of both sides gives x+3 = ±2, so the solutions are x=-1 and x=-5.
- 4x^2 – 12x + 9 = 0 This quadratic equation can be solved by completing the square: 4(x-3/2)^2 – 9/4 = 0. Rearranging gives (x-3/2)^2 = 9/16, so the solutions are x = 3/2 ± 3/4 = 3/4 or 9/4.
These are some examples of quadratic equations that can be solved using different methods, such as factoring, the quadratic formula, or completing the square.
Quadratic Equations Ex 4.1 Questions
Ex 4.1 Class 10 Maths Question 1.
Check whether the following are quadratic equations:
(i) (x+ 1)2=2(x-3)
(ii) x – 2x = (- 2) (3-x)
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
(iv) (x – 3) (2x + 1) = x (x + 5)
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
(vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 -4x2 -x + 1 = (x-2)3
(i) (x+1)²=2(x-3) can be simplified as x² – 4x + 6 = 0. This is a quadratic equation.
(ii) x – 2x = (-2)(3-x) can be simplified as x² – x – 6 = 0. This is a quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) can be simplified as x² – 2x – 3 = 0. This is a quadratic equation.
(iv) (x – 3)(2x + 1) = x(x + 5) can be simplified as 2x² – 7x – 15 = 0. This is a quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) can be simplified as x² – 4x – 4 = 0. This is a quadratic equation.
(vi) x² + 3x + 1 = (x – 2)² can be simplified as x² – 5x + 5 = 0. This is a quadratic equation.
(vii) (x + 2)³ = 2x(x² – 1) can be simplified as x³ + 6x² + 11x + 4 = 0. This is a cubic equation, not a quadratic equation.
(viii) x³ – 4x² – x + 1 = (x – 2)³ can be simplified as x³ – 7x² + 13x – 7 = 0. This is a cubic equation, not a quadratic equation.
Therefore, out of the given equations, (i) to (v) are quadratic equations and (vi) to (viii) are not quadratic equations.
Ex 4.1 Class 10 Maths Question 2.
Let the breadth of the rectangular plot be x meters. Then, its length will be (2x+1) meters (as per the given condition).
The area of a rectangle is given by the formula:
Area = Length x Breadth
Substituting the values of length and breadth in terms of x, we get:
Area = (2x+1) x x = 2x² + x
According to the problem, the area of the rectangular plot is 528 m². Therefore,
2x² + x = 528
This is a quadratic equation in standard form. We can simplify it further to get the solution:
2x² + x – 528 = 0
We can now solve this quadratic equation to find the values of x (breadth) and (2x+1) (length) which will give us the dimensions of the rectangular plot.
According to the problem, the product of these two integers is 306. Therefore,
x(x+1) = 306
Expanding the left-hand side, we get:
x² + x – 306 = 0
This is a quadratic equation in standard form. We can solve this quadratic equation to find the value of x, which will give us the first of the two consecutive positive integers. Once we know the value of x, we can find the next integer by adding 1 to it.
We can solve this quadratic equation by factorization or by using the quadratic formula. Let’s use factorization to solve it:
x² + x – 306 = 0
(x+18)(x-17) = 0
Therefore, x = -18 or x = 17. Since we are looking for positive integers, we reject the negative solution.
Hence, the first of the two consecutive positive integers is 17, and the next integer is 18.
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
How To Stop Message Blocking Is Active error Notification
Let’s assume Rohan’s current age to be x years.
According to the question, Rohan’s mother is 26 years older than him. Therefore, his mother’s current age would be x + 26 years.
After three years, Rohan’s age will be x + 3 years, and his mother’s age will be (x + 26) + 3 = x + 29 years.
The product of their ages (in years) 3 years from now will be 360. So, we can write:
(x + 3) (x + 29) = 360
Expanding the above equation, we get:
x² + 32x – 297 = 0
This is a quadratic equation in standard form, where the coefficient of x² is 1, the coefficient of x is 32, and the constant term is -297.
We can solve this quadratic equation using the quadratic formula:
x = [-b ± sqrt(b² – 4ac)] / 2a
where a = 1, b = 32, and c = -297
Substituting these values in the quadratic formula, we get:
x = [-32 ± sqrt(32² – 4(1)(-297))] / 2(1)
x = [-32 ± sqrt(32² + 42971)] / 2
x = [-32 ± sqrt(10976)] / 2
x = [-32 ± 104] / 2
x = 36 or x = -68
Since Rohan’s age cannot be negative, the only valid solution is:
Rohan’s present age is 36 years.
Let’s assume the speed of the train to be x km/h.
According to the question, the train travels a distance of 480 km at a uniform speed. So, the time taken by the train to cover this distance would be:
time taken = distance / speed = 480 / x
Now, if the speed had been 8 km/h less, then the new speed of the train would be (x – 8) km/h. And the time taken to cover the same distance of 480 km would be:
time taken = distance / speed = 480 / (x – 8)
The question states that if the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Therefore, we can write:
480 / x – 480 / (x – 8) = 3
Simplifying the above equation, we get:
480[(x – 8) – x] / x(x – 8) = 3
480(-8) / [x(x – 8)] = 3
-3840 = 3x² – 24x
3x² – 24x – 3840 = 0
This is a quadratic equation in standard form, where the coefficient of x² is 3, the coefficient of x is -24, and the constant term is -3840.
We can solve this quadratic equation using the quadratic formula:
x = [-b ± sqrt(b² – 4ac)] / 2a
where a = 3, b = -24, and c = -3840
Substituting these values in the quadratic formula, we get:
x = [24 ± sqrt(24² – 4(3)(-3840))] / 2(3)
x = [24 ± sqrt(24² + 433840)] / 6
x = [24 ± sqrt(23328)] / 6
x = [24 ± 152] / 6
x = 29.33 or x = -21.33
Since the speed of the train cannot be negative, the only valid solution is:
The speed of the train is approximately 29.33 km/h (rounded to two decimal places).
Quadratic Equation
A quadratic equation is a second-degree polynomial equation of the form ax² + bx + c = 0, where a, b, and c are constants (coefficients) and x is the variable.
The highest power of x in a quadratic equation is 2, and it can be written in the standard form as:
ax² + bx + c = 0
where a ≠ 0.
To solve a quadratic equation, we usually use the quadratic formula, which states that the solutions (roots) of a quadratic equation can be found using the formula:
x = [-b ± sqrt(b² – 4ac)] / 2a
where the discriminant (b² – 4ac) determines the nature of the solutions. If the discriminant is positive, the quadratic equation has two distinct real roots. If the discriminant is zero, the quadratic equation has one real root (a double root). If the discriminant is negative, the quadratic equation has two complex conjugate roots.
We can also solve quadratic equations by factoring or completing the square, but the quadratic formula is the most general method that can be used to solve any quadratic equation.
Zero(es)/Root(s) of Quadratic Equation
The zeros or roots of a quadratic equation are the values of x that satisfy the equation and make it equal to zero. In other words, the roots of a quadratic equation are the values of x for which the parabolic curve given by the equation intersects the x-axis.
The quadratic equation in standard form, ax² + bx + c = 0, can be solved to find its roots using the quadratic formula:
x = [-b ± sqrt(b² – 4ac)] / 2a
If the discriminant (b² – 4ac) is positive, the quadratic equation has two real roots, which are given by the above formula. If the discriminant is zero, the quadratic equation has one real root, which is:
x = -b / 2a
If the discriminant is negative, the quadratic equation has two complex roots (also known as imaginary roots), which are given by:
x = [-b ± i sqrt(-b² + 4ac)] / 2a
where i is the imaginary unit (i² = -1).
It is important to note that the roots of a quadratic equation can also be found by factoring or completing the square, but the quadratic formula is the most general method that can be used to find the roots of any quadratic equation.
Relation Between Zeroes and Co-efficient of a Quadratic Equation
The relationship between the zeros (roots) and the coefficients of a quadratic equation is given by Vieta’s formulas. Vieta’s formulas provide a relationship between the coefficients and the roots of a quadratic equation, as well as higher degree polynomials.
For a quadratic equation in standard form, ax² + bx + c = 0, the sum and product of the roots are given by:
sum of roots = -b / a
product of roots = c / a
This means that the sum of the roots is equal to the negative coefficient of x divided by the coefficient of x², and the product of the roots is equal to the constant term divided by the coefficient of x².
For example, consider the quadratic equation 2x² – 5x + 3 = 0. The coefficient of x² is 2, the coefficient of x is -5, and the constant term is 3. Using Vieta’s formulas, we can find the sum and product of the roots as:
sum of roots = -(-5) / 2 = 5/2
product of roots = 3 / 2 = 1.5
Therefore, the sum of the roots is 5/2 and the product of the roots is 1.5.
Vieta’s formulas can also be extended to higher degree polynomials. For example, for a cubic equation ax³ + bx² + cx + d = 0, the sum, product, and pairwise product of the roots can be expressed in terms of the coefficients a, b, c, and d.
Methods of Solving Quadratic Equation
There are several methods of solving quadratic equations, including:
- Factorization: If the quadratic equation can be factored, we can use the zero-product property to solve for the roots. For example, the quadratic equation x² – 5x + 6 = 0 can be factored as (x – 2)(x – 3) = 0, which means that the roots are x = 2 and x = 3.
- Completing the square: We can complete the square to rewrite the quadratic equation in vertex form, which makes it easier to find the roots. For example, the quadratic equation x² + 6x + 5 = 0 can be rewritten as (x + 3)² – 4 = 0. Solving for x, we get (x + 3)² = 4, which means that x + 3 = ±2. Therefore, the roots are x = -1 and x = -5.
- Quadratic formula: The quadratic formula can be used to find the roots of any quadratic equation in standard form. The quadratic formula is:
x = (-b ± sqrt(b² – 4ac)) / 2a
For example, the quadratic equation 2x² + 5x – 3 = 0 can be solved using the quadratic formula as:
x = (-5 ± sqrt(5² – 4(2)(-3))) / 2(2)
x = (-5 ± sqrt(49)) / 4
x = (-5 ± 7) / 4
Therefore, the roots are x = -3/2 and x = 1.
- Graphical method: We can plot the quadratic function and find the x-intercepts (where the graph intersects the x-axis) to find the roots. This method is less precise than the other methods, but it can be useful for visualizing the roots and estimating their values.
These are some of the common methods for solving quadratic equations. Depending on the specific equation and its properties, some methods may be more efficient or easier to use than others.
Methods of Factorization
There are several methods of factorization, including:
- Common factor: If a polynomial has a common factor, we can factor it out. For example, the polynomial 6x² + 9x can be factored as 3x(2x + 3).
- Difference of squares: If a polynomial is in the form a² – b², we can factor it as (a + b)(a – b). For example, the polynomial x² – 4 can be factored as (x + 2)(x – 2).
- Perfect square trinomial: If a polynomial is in the form a² + 2ab + b² or a² – 2ab + b², we can factor it as (a + b)² or (a – b)², respectively. For example, the polynomial x² + 6x + 9 can be factored as (x + 3)².
- Grouping: If a polynomial has four or more terms, we can group them in pairs and factor out a common factor from each pair. For example, the polynomial 3x² – 4x – 3 can be factored as (3x + 3)(x – 1) or (x – 1)(3x – 3).
- Sum or difference of cubes: If a polynomial is in the form a³ + b³ or a³ – b³, we can factor it as (a + b)(a² – ab + b²) or (a – b)(a² + ab + b²), respectively. For example, the polynomial x³ + 8 can be factored as (x + 2)(x² – 2x + 4).
- Trial and error: If other methods do not work, we can try different factor pairs until we find the correct ones. This method can be time-consuming, but it can be useful for polynomials that do not fit into any of the other factorization methods.
These are some of the common methods of factorization. The choice of method depends on the specific polynomial and its properties. Sometimes, a combination of methods may be needed to factorize a polynomial.
Method of Completing the Square
y = a(x – h)² + k
where a, h, and k are constants, and (h, k) is the vertex of the parabola.
Here are the steps for completing the square:
- Rewrite the quadratic expression in the form ax² + bx + c = 0.
- Divide both sides of the equation by a (the coefficient of x²), so that the leading coefficient is 1.
- Move the constant term to the right side of the equation, so that the left side is a perfect square trinomial.
- Add and subtract (b/2a)² to the left side of the equation to make it a perfect square trinomial.
- Factor the perfect square trinomial, and simplify the right side of the equation.
- Take the square root of both sides of the equation, and solve for x.
Here is an example:
Solve the quadratic equation x² + 4x – 5 = 0 by completing the square.
- Rewrite the quadratic expression in the form ax² + bx + c = 0: x² + 4x – 5 = 0.
- Divide both sides of the equation by 1: x² + 4x – 5 = 0.
- Move the constant term to the right side of the equation: x² + 4x = 5.
- Add and subtract (4/2*1)² = 4 to the left side of the equation: x² + 4x + 4 – 4 = 5.
- Factor the perfect square trinomial: (x + 2)² – 9 = 5.
- Simplify the right side of the equation: (x + 2)² = 14.
- Take the square root of both sides of the equation: x + 2 = ±√14.
- Solve for x: x = -2 ± √14.
Therefore, the solutions of the quadratic equation x² + 4x – 5 = 0 are x = -2 + √14 and x = -2 – √14.
Quadratic Formula
The quadratic formula is a formula that can be used to find the solutions of a quadratic equation of the form ax² + bx + c = 0, where a, b, and c are constants, and a is not equal to zero. The quadratic formula is given by:
x = (-b ± √(b² – 4ac)) / 2a
The two solutions of the quadratic equation are obtained by substituting the plus and minus signs separately in the formula. The term inside the square root is called the discriminant, and it determines the nature of the solutions of the quadratic equation.
If the discriminant is positive, then the quadratic equation has two distinct real solutions. If the discriminant is zero, then the quadratic equation has one real solution with a multiplicity of two. If the discriminant is negative, then the quadratic equation has two complex conjugate solutions.
Here is an example of how to use the quadratic formula to solve a quadratic equation:
Solve the quadratic equation 2x² – 3x – 5 = 0 using the quadratic formula.
First, we identify the values of a, b, and c from the given quadratic equation:
a = 2 b = -3 c = -5
Substituting these values into the quadratic formula, we get:
x = (-(-3) ± √((-3)² – 4(2)(-5))) / 2(2)
Simplifying this expression, we get:
x = (3 ± √49) / 4
x = (3 ± 7) / 4
Therefore, the solutions of the quadratic equation 2x² – 3x – 5 = 0 are x = -1 and x = 5/2.
The nature of the roots of a quadratic equation ax² + bx + c = 0 is determined by the discriminant, which is given by the expression b² – 4ac. The discriminant helps us to determine whether the roots of the quadratic equation are real or complex, and whether they are distinct or equal. Here are the different cases:
- If the discriminant is positive (b² – 4ac > 0), then the quadratic equation has two distinct real roots.
- If the discriminant is zero (b² – 4ac = 0), then the quadratic equation has one real root with a multiplicity of 2.
- If the discriminant is negative (b² – 4ac < 0), then the quadratic equation has two complex conjugate roots.
We can use the discriminant to classify the nature of the roots of the quadratic equation without actually finding the roots. Here are some examples:
- Consider the quadratic equation x² + 2x + 1 = 0. Here, a = 1, b = 2, and c = 1. The discriminant is given by b² – 4ac = 2² – 4(1)(1) = 0. Since the discriminant is zero, the quadratic equation has one real root with a multiplicity of 2. In this case, the quadratic equation factors as (x + 1)² = 0, so the only root is x = -1.
- Consider the quadratic equation x² + x + 1 = 0. Here, a = 1, b = 1, and c = 1. The discriminant is given by b² – 4ac = 1² – 4(1)(1) = -3. Since the discriminant is negative, the quadratic equation has two complex conjugate roots. In this case, the roots are given by x = (-1 ± i√3)/2, where i is the imaginary unit.
- Consider the quadratic equation x² + 4x + 5 = 0. Here, a = 1, b = 4, and c = 5. The discriminant is given by b² – 4ac = 4² – 4(1)(5) = -4. Since the discriminant is negative, the quadratic equation has two complex conjugate roots. In this case, the roots are given by x = (-4 ± 2i)/2 = -2 ± i.
FAQs About Class 10 Math Quadratic Equation solution
- What is a quadratic equation? A quadratic equation is a polynomial equation of degree two, in which the highest power of the variable is 2.
- What is the standard form of a quadratic equation? The standard form of a quadratic equation is ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.
- How many solutions does a quadratic equation have? A quadratic equation can have two solutions, one solution, or no real solutions, depending on the discriminant (b^2 – 4ac) of the equation.
- How can I solve a quadratic equation? There are several methods for solving a quadratic equation, including factoring, completing the square, and using the quadratic formula.
- What is the discriminant of a quadratic equation? The discriminant of a quadratic equation is the expression b^2 – 4ac, which appears under the square root in the quadratic formula. It can be used to determine the nature and number of solutions of the equation.
- What is the quadratic formula? The quadratic formula is x = (-b ± sqrt(b^2 – 4ac)) / 2a, which gives the solutions of a quadratic equation of the form ax^2 + bx + c = 0.
- What is the vertex of a parabola? The vertex of a parabola is the point where the parabola intersects its axis of symmetry. It is a turning point of the parabola and has the coordinates (-b/2a, f(-b/2a)).
- What are the properties of the solutions of a quadratic equation? The solutions of a quadratic equation have several properties, such as being symmetric with respect to the vertex of the parabola, and having a product equal to the constant term divided by the leading coefficient.
(FAQs) about NCERT Solutions for Class 10 Maths Chapter 4:
- What topics are covered in NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equations? NCERT Solutions for Class 10 Maths Chapter 4 cover topics such as the standard form of a quadratic equation, methods of solving quadratic equations, discriminant, nature of roots, and quadratic equations in real-life situations.
- What is the importance of NCERT Solutions for Class 10 Maths Chapter 4? NCERT Solutions for Class 10 Maths Chapter 4 is important as it lays the foundation for understanding higher-level mathematical concepts such as calculus, algebra, and geometry. It helps students to understand the properties and behavior of quadratic equations, which are used in various fields such as physics, engineering, and finance.
- Are NCERT Solutions for Class 10 Maths Chapter 4 useful for board exams? Yes, NCERT Solutions for Class 10 Maths Chapter 4 is very useful for board exams as it covers all the important concepts and topics related to quadratic equations. The solutions provided in the NCERT textbook are designed by experts and are in line with the CBSE syllabus, making them very helpful for exam preparation.
- How can I access NCERT Solutions for Class 10 Maths Chapter 4? NCERT Solutions for Class 10 Maths Chapter 4 can be accessed online through various websites or can be downloaded as a PDF file. Additionally, students can refer to the NCERT textbook and practice the exercises given at the end of each chapter.
- Can NCERT Solutions for Class 10 Maths Chapter 4 help improve my grades? Yes, NCERT Solutions for Class 10 Maths Chapter 4 can definitely help improve your grades as it provides a thorough understanding of the concepts and formulas related to quadratic equations. By practicing the exercises and solving the problems, students can improve their problem-solving skills and score better in exams.