Question

One of the roots of equation x2 + mx - 5 = 0 is 2 ; find m. *

1. -2

2. 2

3. 1/2

4. -1/2

​

Answer 1

Given equation

$x^{2} +mx-5=0$

2 is the zero of the polynomial.

Substituting x = 2, in the equation, we get,

$\implies (2)^{2} +m(2)-5=0$

$\implies 4+2m-5=0$

$\implies 2m-1=0$

$\implies 2m=1$

$\implies m = \dfrac{1}{2}$

$\boxed{\boxed{\bold{Therefore, \ the \ option \ 3, \ m \ = \ \frac{1}{2} \ is \ the \ answer}}}}}}}}}$

                                                           

Answer 2

$\huge\sf\underline\purple{Solution}$

Given QE x² + mx -5 =0

And One of root is 2

Hence 2 is a root of QE So ,Substuite in Given equation to get the value of m

${x² + mx -5 =0}$

${(2)² +2m - 5 =0}$

${4 + 2m -5 =0}$

${2m - 1 =0}$

${2m = 1}$

${m}$=$\frac{1}{2}$

Hope u helpful