Class 10 Solution of a Quadratic Equation by Factorisation

### Topic to be coverd

♦ Solution of a Quadratic Equation by Factorisation

### Solution of a Quadratic Equation by Factorisation

color{green} ✍️ In general, a real number α is called color(green)("a root of the quadratic equation") color(blue)(ax^2 + bx + c = 0, a ≠ 0)

if color(red)(a α^2 + bα + c = 0.) We also say that x = α is a solution of the quadratic equation, or that color(blue)("α satisfies the quadratic equation.")

color(green)(Note ) that the zeroes of the quadratic polynomial ax^2 + bx + c and the roots of the quadratic equation ax^2 + bx + c = 0 are the same.

color(red)(=>"Consider the quadratic equation") color(red)(2x^2 – 3x + 1 = 0.)

If we replace x by 1 on the LHS of this equation,

we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation.

We say that 1 is a root of the quadratic equation 2x^2 – 3x + 1 = 0.

This also means that 1 is a zero of the quadratic polynomial 2x^2 – 3x + 1.

You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes.

color{green} ✍️ color(green)("So, any quadratic equation can have atmost two roots.")

color{green} ✍️ You have learnt in Class IX, how to factorise quadratic polynomials by splitting their middle terms. We shall use this knowledge for finding the roots of a quadratic equation.
Q 3169767615

Find the roots of the equation 2x^2 – 5x + 3 = 0, by factorisation.
Class 10 Chapter 4 Example 3
Solution:

Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x^2 = (2x^2) × 3].
So, 2x^2 – 5x + 3 = 2x^2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)
Now, 2x^2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x^2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives x = 3/2 and x-1 = 0 gives x = 1.

so x = 3/2 and x = 1 are the solutions of the eqaution.

In other words , 1 and 3/2 are the roots of the equation 2x^2-5x+3 = 0.

Verify that these are the roots of the given equation.

Note that we have found the roots of 2x^2-5x+3 = 0 by factorising 2x^2-5x+3 into two linear factors and equating each factor to zero.
Q 3189767617

Find the roots of the quadratic equation 6x^2 – x – 2 = 0.
Class 10 Chapter 4 Example 4
Solution:

We have
6x^2 – x – 2 = 6x^2 + 3x – 4x – 2
= 3x (2x + 1) – 2 (2x + 1)
= (3x – 2)(2x + 1)
The roots of 6x^2 – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0
Therefore, 3x – 2 = 0 or 2x + 1 = 0,

i.e. x = 2/3 or x = -1/2

Therefore, the roots of 6x^2 – x – 2 = 0 are 2/3 and -1/2.

We verify the roots, by checking that 2/3 and -1/2 satisfy 6x^2-x-2 = 0
Q 3129867711

Find the roots of the quadratic equation 3x^2 - 2 sqrt6x + 2 = 0 .
Class 10 Chapter 4 Example 5
Solution:

3x^2 - 2 sqrt(6x) + 2 = 3x^2-sqrt6x - sqrt6x +2

 = sqrt3x( sqrt3x-sqrt2) - sqrt2 ( sqrt3x-sqrt2)

= ( sqrt3x - sqrt2) ( sqrt3x-sqrt2)

So, the roots of the equation are the values of x for which

( sqrt3x-sqrt2) ( sqrt3x-sqrt2) = 0

Now sqrt3x-sqrt2 = 0 for x = sqrt(2/3)

So, this root is repeated twice, one for each repeated factor sqrt3x-sqrt2.

Therefore, the roots of 3x^2-2 sqrt6x+2 = 0 are sqrt(2/3) , sqrt(2/3)
Q 3189067817

Find the dimensions of the prayer hall discussed in a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall ?
Class 10 Chapter 4 Example 6
Solution:

Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. We can depict
this information pictorially as shown in Fig.
Now, area of the hall = (2x + 1). x m^2 = (2x^2 + x) m^2
So, 2x^2 + x = 300 (Given)
Therefore, 2x^2 + x – 300 = 0

Applying the factorisation method, we write this equation as

2x^2 – 24x + 25x – 300 = 0
2x (x – 12) + 25 (x – 12) = 0
i.e., (x – 12)(2x + 25) = 0
So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.