Question

Find the roots of the following quadratic equations by factorisation:
√7 y2 – 6y — 13√7 = 0

Answer 1

Question:

Find the roots of the following quadratic equation by factorisation : √7y² - 6y - 13√7 = 0

Answer:

x = 13√7/7 , -√7

Note:

• The possible values of unknown (variable) for which the equation is satisfied are called its solutions or roots .

• If x = a is a solution of any equation in x , then it must satisfy the given equation otherwise it's not a solution (root) of the equation.

Solution:

The given quadratic equation is :

√7y² - 6y - 13√7 = 0 .

Now ,

Splitting the middle term of the given quadratic equation , we have ;

=> √7y² - 6y - 13√7 = 0

=> √7y² - 13y + 7y - 13√7 = 0

=> y(√7y - 13) + √7(√7y - 13) = 0

=> (√7y - 13)•(y + √7) = 0

Case1 : √7y - 13 = 0

=> √7y - 13 = 0

=> √7y = 13

=> y = 13/√7

=> y = 13√7/7

Case2 : y + √7 = 0

=> y + √7 = 0

=> y = - √7

Hence,

Hence,The required roots of the given quadratic equation are : y = 13√7/7 , -√7

Answer 2

Answer:

$\large\boxed{\sf{-\sqrt{7}\;\;and\;\;\dfrac{13}{\sqrt{7}}}}$

Step-by-step explanation:

Given a quadratic equation such that,

$\sqrt{7} {y}^{2} - 6y - 13 \sqrt{7} = 0$

We have to find the roots.

Also, we have to find the roots by factorisation method.

This type of equation can be factorised by middle term splitting method.

Therefore, we will get,

$= > \sqrt{7} {y}^{2} + 7y - 13y - 13 \sqrt{7} = 0$

Now, taking the common terms, we get,

$= > \sqrt{7} y(y + \sqrt{7} ) -1 3(y + \sqrt{7} ) \\ \\ = > (y + \sqrt{7})( \sqrt{7} y - 13) = 0$

Now, we have two cases.

Case I

When y + √7 = 0

=> y = -√7

Case II

When √7y - 13 = 0

=> √7y = 13

=> y = 13/√7

Hence, the roots are $\bold{-\sqrt{7}\;\;and\;\;\dfrac{13}{\sqrt{7}}}$