# Given that x=2 is a root of the equation x^2 +x +m =0, find the other root and the value of m.

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### 3 Answers

Given that x = 2 is a root of the equation x^2 +x +m =0. Let the other root be x = p.

This is a quadratic equation having two roots.

Hence (x-2)(x-p) = x^2 +x +m

rArr x^2-x(p+2) +2p = x^2 +x +m

Comparing the coefficients of x and the constant term, we get -(p+2) = 1

rArr p =-3

and also, m = 2p = 2*(-3)=-6.

**Therefore, the other root of the equation x^2 +x +m =0 is x = -3, and the value of m is -6**.

**Sources:**

x^2 +x +m =0

2^2+2+m=0

4+2+m=0

6+m=0

m=-6

`x^2+x-6=0 `

factors of -6 that add to 1 are 3 and -2

`x^2+3x-2x-6 `

`x(x+3)-2(x+3)`

`(x+3)(x-2)`

x=-3 x=2

the other root is -3

Given: x=2 is a root of the equation x^2+x+m=0.

Thus, substituting the value of x, in this equation, we get,

2^2+2+m=0

or, 6+m=0

**Therefore, m=-6.**

Now, substituting the value of m in the given equation, we will factorize it to get the other root.

x^2+x-6=0

or, x^2+3x-2x-6=0

or, x(x+3)-2(x+3)=0

or, (x+3)(x-2)=0

Either, Or,

x+3=0 x-2=0

Thus, x=-3. Thus, x=2.

**Therefore, the other root of the equation is -3.**