Question

If 1/2 is aroot of the equation x^2+kx-5/4=0, then the value of k is
a. 2
b. -2
c. 1/4
d. 1/2​

Answer 1

$\bf{\underline{\underline{Question:-}}}$

If 1/2 is a root of the equation x^2+kx-5/4=0, then the value of k is

a. 2

b. -2

c. 1/4

d. 1/2

$\bf{\underline{\underline{Solution:-}}}$

$\sf → x^2+kx-\frac{5}{4}=0$

• Substituting 1/2 on the place of X in the polynomial.

$\sf → (\frac{1}{2})^2+k×\frac{1}{2}-\frac{5}{4}=0$

$\sf → \frac{1}{4}+\frac{1}{2}k-\frac{5}{4}=0$

• Now taking L.C.M of 4 , 2 , 4 = 4

$\sf → \frac{1+2k-5}{4}=0$

$\sf → \frac{-4+2k}{4}=0$

$\sf → -4+2k=0×4$

$\sf → -4+2k=0$

$\sf → 2k=4$

$\sf → k=\dfrac{\cancel{4}}{\cancel{2}}$

$\sf → k = 2$

$\bf{\underline{\underline{Hence:-}}}$

• The required value of k is 2

Answer 2

Answer:

2

Step-by-step explanation:

x^2+kx - 5/4 = 0

1/2^2 + k*1/2 - 5/4 =0

1/4 +k/2 - 5/4 = 0

(1 +2k - 5)/4 = 0

-4 + 2k = 0*4

-4 +2k = 0

2k = 4

k = 4/2

k = 2