Question

if root 2 is a root of quadratic equation kx^2 +root 2 -4 find the value of k​

Answer 1

Correct Question:

$\sf{If \ \sqrt2 \ is \ a \ root \ of \ quadratic \ equation} \\ \sf{kx^{2}+\sqrt2x-4=0 \ find \ the \ value \ of \ k.}$

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Answer:

$\sf{The \ value \ of \ k \ is \ 1.}$

Given:

$\sf{The \ given \ quadratic \ equation \ is} \\ \sf{kx^{2}+\sqrt2x-4=0, \ \sqrt2 \ is \ root \ of \ equation.}$

To find:

$\sf{The \ value \ of \ k.}$

Solution:

$\sf{\leadsto{kx^{2}+\sqrt2x-4=0}} \\ \\ \sf{Substitute \ x=\sqrt2, \ we \ get} \\ \\ \sf{2k+2-4=0} \\ \\ \sf{\therefore{2k=2}} \\ \\ \sf{\therefore{k=1}} \\ \\ \purple{\tt{\therefore{The \ value \ of \ k \ is \ 1.}}}$

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Verification:

$\sf{\leadsto{kx^{2}+\sqrt2x-4=0}} \\ \\ \sf{Substitute \ k=1, \ we \ get} \\ \\ \sf{x^{2}+\sqrt2x-4=0} \\ \\ \sf{\therefore{x^{2}+2\sqrt2x-\sqrt2x-4=0}} \\ \\ \sf{\therefore{x(x+2\sqrt2x)-\sqrt2(x+2\sqrt2)=0}} \\ \\ \sf{\therefore{(x+2\sqrt2x)(x-\sqrt2)=0}} \\ \\ \sf{\therefore{x=-2\sqrt2 \ or \ \sqrt2}} \\ \\ \sf{Since, \ one \ of \ the \ root \ is \ \sqrt2} \\ \\ \sf{Hence, \ verified.}$