if 1/2 is a root of the equation 2x²-kx+3=0,then find the value of k
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if 1/2 is a root of the equation 2x²-kx+3=0,then find the value of k
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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
[tex]\boxed {\underline {\mathbb {FINAL\:ANSWER:-}}}[/tex]
[tex]\boxed {k=-7}[/tex]
[tex]\boxed {\underline {\mathbb {GIVEN:-}}}[/tex]
[tex]\frac{1}{2}[/tex] is one of the root of [tex]2x^{2}-kx+3=0[/tex]
[tex]\boxed {\underline {\mathbb {TO\:FIND:-}}}[/tex]
value of k
[tex]\boxed {\underline {\mathbb {SOLUTION:-}}}[/tex]
It’s given that [tex]\frac{1}{2}[/tex] is one of the root of [tex]2x^{2}-kx+3=0[/tex]
Means [tex]x=\frac{1}{2} \implies 2x=1 \implies 2x-1=0 \implies (2x-1)[/tex]
As [tex]\frac{1}{2}[/tex] is root of equation which means that ii will give 0
Now as we know x value so let’s put in equation
[tex]2x^{2}-kx+3=0\\2(\frac{1}{2})^{2}-k(\frac{1}{2})+3=0[/tex]
[tex]2(\frac{1}{4}) - \frac{k}{2}+3=0[/tex] [← now 2 will cut 4 gives [tex]\frac{1}{2}[/tex]]
[tex]\frac{1}{2} -\frac{k}{2}+3=0\\ \frac{1 \times 1 + k \times 1 + 6}{2}=0\\\frac{1+k+6}{2}=0\\[/tex]
[tex]\frac{7+k}{2}=0[/tex] [ ← bringing 2 to RHS get multiplied to 0 gives 0]
[tex]7+k=0\\K=-7[/tex]
Hence k = -7
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