Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If \frac{1}{2} is a root of the equation x^{2}+kx-\frac{5}{4}=0 , then the value of k is (A)2\\ (B)-2\\ (C)\frac{1}{4}\\ (D)\frac{1}{2}

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

(A)2\\ (B)-2\\ (C)\frac{1}{4}\\ (D)\frac{1}{2}

Answers (1)

(A) 2

Solution

As we know that if \frac{1}{2}  is a root of the equation x^{2}+kx-\frac{5}{4}=0  then x=\frac{1}{2}  should satisfy its equation.   

Put       x=\frac{1}{2}

 \\\left (\frac{1}{2} \right )^{2}+k\left (\frac{1}{2} \right )-\frac{5}{4}=0\\ \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0\\ \frac{1+2k-5}{4}=0\\ 2k-4=0\\ 2k=4\\ k=\frac{4}{2}=2\\ k=2

Hence value of k is 2.            

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question