Question

if x=√2is the solution of the equation kx2+√2x-4=0,then find the value of K​

Answer 1

Hey Pretty Stranger!

$\sf \: k {x}^{2} + \sqrt{2} x - 4 = 0$

Put the value of x = √2

$\longrightarrow \sf \: k( \sqrt{2 }) ^{2} + \sqrt{2} ( \sqrt{2}) + 4 = 0$

$\longrightarrow \sf \: 2k + 2 - 4 = 0$

$\longrightarrow \sf \: 2k - 2= 0$

$\longrightarrow \sf \: 2k =0 + 2$

$\longrightarrow \sf \: 2k =2$

$\longrightarrow \sf \: k = \dfrac{2}{2}$

$\longrightarrow \large \boxed{ \rm \: k =1}$

Answer 2

Answer:

answer is K=1

see the image

you can check by putting the value of K=1 in equation

1×√2×√2+√2×√2-4

1×2+2-4

4-4=0

RHS=0

so it is right