Question

please solve this question ​

Answer 1

Answer:

x =1 ,x=5

Step-by-step explanation:

hope it is helpful

Answer 2

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Given :

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

To Find :

Factorise and hence find roots .

Above equation is in the form of a quadratic equation e.g,$\bold{\red{a {x}^{2} + bx + c}}$

So ,we simply factorise it and find roots.

How to factorise?

step1: Multiply the constant term with coefficient of x square.suppose we obtain 'z' after multiplying

step 2:Think of a number whose sum equal to the coefficient of middle term and product makes the 'z'

step 3: now split the middle term by factors of that number which you think.

step 4:At last take common and compare with equal to 0 we obtain the roots .

Come to the question:

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

$\sf=>5 \times 1(coefficient \: of \: {x}^{2} )=5$

5x+x=6x

Now ,

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

$\sf= > x(x + 5) + 1(x + 5) = 0$

$\sf= > (x + 1)(x + 5) = 0$

$\sf x + 1 = 0$

$\bold{x = - 1}$

$\sf x + 5 = 0$

$\bold{x = - 5}$

$\therefore$Roots are x=-5 and x=-1.