Question

if 1/2 is a root of the equation 2x²-kx+3=0,then find the value of k​

Answer 1

Step-by-step explanation:

2x²-kx+3= 0

= 2×(1/2)²-k(1/2)+3= 0

= 2× 1/4 - k/2 +3= 0

= 1/2-k/2+3= 0

= 1-k+6=0

= -k+7=0

= -k= -7

k= 7

Answer 2

$\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}$

$\boxed {k=-7}$

$\boxed {\underline {\mathbb {GIVEN:-}}}$

$\frac{1}{2}$ is one of the root of $2x^{2}-kx+3=0$

$\boxed {\underline {\mathbb {TO\:FIND:-}}}$

value of k​

$\boxed {\underline {\mathbb {SOLUTION:-}}}$

It’s given that $\frac{1}{2}$ is one of the root of $2x^{2}-kx+3=0$

Means $x=\frac{1}{2} \implies 2x=1 \implies 2x-1=0 \implies (2x-1)$

As $\frac{1}{2}$ is root of equation which means that ii will give 0  

Now as we know x value so let’s put in equation

$2x^{2}-kx+3=0\\2(\frac{1}{2})^{2}-k(\frac{1}{2})+3=0$

$2(\frac{1}{4}) - \frac{k}{2}+3=0$  [← now 2 will cut 4 gives $\frac{1}{2}$]

$\frac{1}{2} -\frac{k}{2}+3=0\\ \frac{1 \times 1 + k \times 1 + 6}{2}=0\\\frac{1+k+6}{2}=0$

$\frac{7+k}{2}=0$  [ ← bringing 2 to RHS get multiplied to 0 gives 0]

$7+k=0\\K=-7$

Hence k = -7

⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔