Question

(g) (x - 1)2 – 5(x - 1)- 6 = 0 by factorisation​

Answer 1

Answer:

If $(x - 1)^2 - 5(x - 1)- 6 = 0$ then$x^2 - 7x + 3 = 0$.

Step-by-step explanation:

Formula:

$(a - b)^2 = a^2 - 2ab + b^2$

Given:

$(x - 1)^2 - 5(x - 1)- 6 = 0\\(x^2 - 2x + 2^2 ) - (5x - 5) - 6 =0\\x^2 - 2x + 4 - 5x + 5 - 6 =0\\x^2 - 2x - 5x + 4 + 5 - 6 =0$

$x^2 -7x + 3 =0$

Answer 2

In context to question asked,

We have to solve the given equation by the method of factorization.

As per question,

it is given that,

$(x - 1)^2 - 5(x - 1)- 6 = 0$

So first we will solve the $(x-1)^2$ by using the identities $(a-b)^2=a^2+b^2-2ab$

Hence, we will get,  

$(x - 1)^2 - 5(x - 1)- 6 = 0\\(x^2 - 2x + 2^2 ) - (5x - 5) - 6 =0\\x^2 - 2x + 4 - 5x + 5 - 6 =0\\x^2 - 2x - 5x + 4 + 5 - 6 =0\\x^2 -7x + 3 =0$