NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations (Ex 4.1) Exercise 4.1

The history of the solution of quadratic equations extends across the world for more than 4000 years as class 10 Math chapter 4 exercise 4.1 will tell you. The earliest known records are Babylonian clay tablets from about 1600 BCE where the diagonal of a unit square is given to five decimal places of accuracy. But, as you will know from class 10 quadratic equation exercise 4.1, the use of algebraic symbols only began in the 15th century. Exercise 4.1 Maths class 10 will also inform you that Euclid, a famous Greek mathematician invented the quadratic equations and its solutions while calculating dimensions such as length; breadth and height via geometry. Later Sridharacharya made the calculations of these quadratic equations by completing the square methodology which became more convenient; better known as now as Sridharacharya Formula of Quadratic Equation.

Class 10 Maths Chapter 4 Exercise 4.1 - Concept of Quadratic Equation

The root of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) is given in NCERT class 10 Maths chapter 4 exercise 4.1 by the following formula:

\[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\]

This formula is known as Sridharacharya formula, as you will know from exercise 4.1 class 10 Maths.

Here, the quantity   b2 −4ac   is called the discriminant of the polynomial.

According to Sridharacharya, as mentioned in NCERT Maths class 10 chapter 4 exercise 4.1, every quadratic equation has two roots and the types depend on the value of this discriminant.

If b2−4ac  > 0, the roots of the quadratic equation are real and distinct in nature.

If b2−4ac  < 0, the roots of the quadratic equation are unreal i.e. has a complex solution.

If b2−4ac   = 0, the roots of the quadratic equation are real and equal.

Example 1:

Examine the nature of the roots of the following quadratic equation.

3x2 + 8x + 4  =  0.

Solution:

The given quadratic equation is in the general form:

ax2 + bx + c  =  0

Then, we have a  =  3, b  =  8 and c  =  4.

Therefore: 

b2 - 4ac  =  82 - 4(3)(4)

b2 - 4ac  =  64 - 48

b2 - 4ac  =  16

Here, b2 - 4ac > 0 and also a perfect square.

So, the roots are real and distinct.

Example 2:

Examine the nature of the roots of the following quadratic equation.

5x2 - 4x + 2  =  0

Solution:

The given quadratic equation is in the general form:

ax2 + bx + c  =  0

Then, we have a  =  5, b  =  -4 and c  =  2.

Therefore:

b2 - 4ac  =  (-4)2 - 4(5)(2)

b2 - 4ac  =  16 - 40

b2 - 4ac  =  -24

Here, b2 - 4ac < 0.

So, the roots are imaginary.

Ex 4.1 Class 10 - Solving Techniques of Quadratic Equations

A quadratic equation may be solved either by factorizing the left side( when the right side is zero) or by completing a square on the left side.

Example 1: Solve  6x2 + 11x - 10 = 0

Solution: 6x2 + 11x - 10 = 0

⇒ 6x2 + 15x - 4x - 10 = 0

⇒ 3x(2x + 5) - 2(2x + 5) = 0

⇒ (2x + 5)(3x - 5) = 0

Therefore, either 2x + 5 = 0 or, 3x - 5 = 0

Hence, x = -5/2 or x = ⅔.

Class 10 Chapter 4 Exercise 4.1: Conversion of Non-Quadratic to Quadratic Equation

Sometimes the equations given are not in quadratic order and hence we need to convert it into a quadratic form to solve it.

Example 1: Solve \[2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}\]

Solution: \[2^{2x + 3} + 2^{x + 3} = 1 + 2^{x}\]

or, \[2^{2x} . 2^{3} + 2^{x} . 2^{3} = 1 + 2^{x}\]

or, \[8 . 2^{2x} + 8 . 2^{x} = 1 + 2^{x}\]…(1)

Now, let \[2^{x} = y\] Then, \[y^{2} = (2^{x})^{2} = 2^{2x}\]

∴ from (1) we get,

\[8y^{2} + 8y = 1 + y\]

or, \[8y^{2} + 7y - 1 = 0\]

or, \[8y^{2} - 8y - y - 1 = 0\]

or, \[8y(y + 1) - 1(y + 1) = 0\]

or, \[(y + 1)(8y - 1) = 0\]

∴ y = -1

or ⅛

If, y = -1, then, \[2^{x} = -1\], which is not possible.

If, y = ⅛, then \[2^{x} = \frac{1}{8}\]

or, \[2^{x} = \frac{1}{2^{3}} = 2^{-3}\]

∴ x = -3.

Example 2: Solve \[\sqrt{\frac{x}{1 - x}} + \sqrt{\frac{1 - x}{x}} = \frac{13}{6}\]

Solution: Let us put \[y = \sqrt{\frac{1}{1-x}}\], then \[\frac{1}{y} = \sqrt{\frac{1 - x}{x}}\]

\[\therefore y + \frac{1}{y} = \frac{13}{6} or, \frac{y^{2} + 1}{y} = \frac{13}{6}\]

⇒ 6y2 + 6 = 13y

⇒ 6y2 - 13y + 6 = 0

⇒ 6y2 - 4y - 9y + 6 = 0

⇒ (3y - 2)(2y - 3) = 0

∴ y = ⅔ or, y = 3/2

If y = ⅔ , then \[\sqrt{\frac{x}{1-x}} = \frac{2}{3}\]

\[\Rightarrow \therefore \frac{x}{1 - x} = \frac{4}{9}\]

⇒ 9x = 4 - 4x

⇒ 13x = 4

∴ x = 4/13

If y = 3/2, then \[\sqrt{\frac{x}{1-x}} = \frac{3}{2}\]

\[\Rightarrow \therefore \frac{x}{1 - x} = \frac{9}{4}\]

⇒ 4x = 9 - 9x

⇒ 13x = 9

∴ x = 9/13

Class 10 Maths Ex 4.1 - Sum and Product of Roots of a Quadratic Equation

If α and β be the roots of the quadratic equation ax2 + bx + c = 0 (a ≠ 0) then,

\[\alpha + \beta = -\frac{b}{a}\] is called the sum of roots and \[\alpha \beta = \frac{c}{a}\] is called products of roots of the quadratic equations.

Example 1: If the roots of the equation x2 + px + 7 = 0 are denoted by α and β and α2 + β2 = 22, find the value of p.

Solution: \[\alpha + \beta = \frac{-p}{1} = (-p)\] and \[\alpha \beta = \frac{7}{1} = 7\]

Now, α2 + β2 = 22 (given)

or, (α + β)2 - 2αβ = 22

or, (-p)2 - 2(7) = 22

or, p2 = 22 + 14 = 36

p = ±6

Class 10 Maths Ch 4 Ex 4.1- Problems based on Quadratic Equation

Class 10th Maths Chapter 4 exercise 4.1 has several problems for you to solve. Following are a few of them.

Example - 1: In a class test, the sum of Rekha’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solution: with the help of class 10 Maths exercise 4.1, let us marks in Maths as xx,
Then as per the problem,

Marks in English would be = 30 − x30 − x.

Now if Rekha got 2 marks less, then Maths marks would = (x − 2)(x − 2)

and if she got 3 marks less in English, then English marks would = (30 − x − 3) = (27 − x)(30 − x − 3) = (27 − x)

As per problem, the product is equal to 210 so

(x − 2)(27 − x) = 120(x − 2)(27 − x) = 120

−x2 + 25x + 54 = 210 − x2 + 25x + 54 = 210

or

x2 − 25 x + 156 = 0 x 2 − 25 x +156 = 0

x2 − 90x + 30x − 2700 = 0 x 2 − 90x + 30x − 2700 = 0

or x = 12 or 13

So Rekha’ marks in Maths is 12 or 13 and the English marks are 18 or 17.

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