Question

*Which of the following can be used to find the roots of a quadratic equation?*

1️⃣ Factorisation method
2️⃣ Completing the square method
3️⃣ Quadratic formula
4️⃣ All of these​

Answer 1

SOLUTION

TO CHOOSE THE CORRECT OPTION

Which of the following can be used to find the roots of a quadratic equation

1. Factorisation method

2. Completing the square method

3. Quadratic formula

4. All of these

EVALUATION

We know that a quadratic equation is of the form

$\sf{a {x}^{2} + bx + c = 0 \: \: \: \: where \: \:a \ne 0 }$

This equation can be solved by the below ways :

1. Factorisation method

Example :

We take a quadratic equation as

$\sf{x(x + 3) - 6 = 12}$

Now

$\sf{x(x + 3) - 6 = 12}$

$\sf{ \implies \: {x}^{2} + 3x - 6 - 12 = 0}$

$\sf{ \implies \: {x}^{2} + 3x - 18 = 0}$

$\sf{ \implies \: {x}^{2} + 6x - 3x - 18 = 0}$

$\sf{ \implies \:x(x + 6) - 3(x + 6) = 0}$

$\sf{ \implies \:(x + 6) (x - 3) = 0}$

$\sf{ \therefore \: \: either \: (x + 6) = 0 \: \: or \: \: (x - 3) = 0 }$

$\sf{ (x + 6) = 0 \: \: \: gives \: \: x = - 6}$

$\sf{ (x - 3) = 0 \: \: \: gives \: \: x = 3}$

∴ The roots are - 6 , 3

2. Completing the square method

Example :

We take the quadratic equation as

$\sf{6 {x}^{2} + 5x + 1 = 0 }$

Since it is a quadratic equation

So it has two roots

$\sf{6 {x}^{2} + 5x + 1 = 0 }$

Dividing both sides by 6 we get

$\displaystyle \: \sf{ {x}^{2} + \frac{5x}{6} + \frac{1}{6} = 0 }$

$\implies \displaystyle \: \sf{ {x}^{2} + 2.x. \frac{5}{12} + { \bigg( \frac{5}{12} \bigg)}^{2} - { \bigg( \frac{5}{12} \bigg)}^{2} + \frac{1}{6} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} - \frac{25}{144} + \frac{1}{6} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} - \frac{1}{144} = 0 }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)}^{2} = \frac{1}{144} }$

$\implies \displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = \pm \: \frac{1}{12} }$

Now

$\displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = \frac{1}{12} } \: \: \: gives$

$\displaystyle \sf{x = \frac{1}{12} - \frac{5}{12} }$

$\implies\displaystyle \sf{x = - \frac{4}{12} }$

$\implies\displaystyle \sf{x = - \frac{1}{3} }$

Again

$\displaystyle \sf{ { \bigg( x + \frac{5}{12} \bigg)} = - \frac{1}{12} } \: \: \: gives$

$\displaystyle \sf{x = - \frac{1}{12} - \frac{5}{12} }$

$\implies\displaystyle \sf{x = - \frac{6}{12} }$

$\implies\displaystyle \sf{x = - \frac{1}{2} }$

Hence the required roots of the quadratic equation are

$\displaystyle \sf{ - \frac{1}{3} \: \: \: and \: \: - \frac{1}{2} }$

3. Quadratic formula

Rule :

For any quadratic equation

$\sf{a {x}^{2} + bx + c = 0 \: \: \: \: where \: \:a \ne 0 }$

The roots are given by

$\displaystyle \: x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac } }{2a}$

FINAL ANSWER

Hence the correct option is

4. All of these

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Answer 2

Answer:

option (4) all the above is correct.