find the value of k for which x=root2 is a solution of the equation kx²+root2x-4=0
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put the value of k in equation
k x [√2]² + √2 x √2 - 4 = 0
2k +2-4 = 0
2k-2 = 0
2k = 2
k = 1
k x [√2]² + √2 x √2 - 4 = 0
2k +2-4 = 0
2k-2 = 0
2k = 2
k = 1
Answered by
6
Hiii,
X= ✓2
p(x) = Kx²+ ✓2x -4
putting the value of X in p(x).
p(✓2) = K× (✓2)² + ✓2 × ✓2 -4
= K× 2 + 2 -4= 2k-2
2K -2 = 0
K = 2/2 = 1..
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X= ✓2
p(x) = Kx²+ ✓2x -4
putting the value of X in p(x).
p(✓2) = K× (✓2)² + ✓2 × ✓2 -4
= K× 2 + 2 -4= 2k-2
2K -2 = 0
K = 2/2 = 1..
I HOPE IT WILL HELP YOU...MARK IT AS BRAINLEIST IF YOU LIKED MY THIS..
THANK YOU..
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