Question

If one root of the quadratic equation kx^2-3x-1=0 is 1/2, then find the value of k.

Answer 1

           

$\huge\boxed{\ \ \ \ \ \ \ \huge\boxed{\ \ \ \ \ \ \ \huge\boxed{\ \ \ \ \ \ \ \bold{k=10}\ \ \ \ \ \ \ }\ \ \ \ \ \ \ }\ \ \ \ \ \ \ }$

   

       

     

$p(x)=kx^2-3x-1=0 \\ \\ \\ p(\frac{1}{2})=k(\frac{1}{2})^2-3(\frac{1}{2})-1=0 \\ \\ p(\frac{1}{2})=k \times \frac{1}{4}-3 \times \frac{1}{2}-1=0 \\ \\ p(\frac{1}{2})=\frac{k}{4}-\frac{3}{2}-1=0 \\ \\ \\ \frac{k}{4}-\frac{3}{2}-1=0 \\ \\ \Rightarrow\ 4(\frac{k}{4}-\frac{3}{2}-1)=4 \times 0 \\ \\ \Rightarrow\ k-6-4=0 \\ \\ \Rightarrow\ k-10=0 \\ \\ \Rightarrow\ k=\bold{10}$

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Answer 2

If one root of the quadratic equation kx^2-3x-1=0 is 1/2, then find the value of k.

Solution :

x = 1/2 is the root of the given quadratic equation.

Therefore, substitute the value of x in the given equation.

.•. k (1/2)^2 - 3 × 1/2 - 1 = 0

k/4 - 3/2 - 1 = 0

k/4 - 1/2 = 0

k/4 = 1/2

2k = 4

k = 2

Hence,

Value of k is 2.

Hope it helps!