Question

If one of the zero of the quadratic polynomial is -4 then find the value of (k-1)^2+kx+1

Answer 1

/* There is a mistake in the question. It should be like this */

$Let \: the \: Quadratic \: polynomial\:$

$p(x) = (k-1)x^{2} +kx+1$

$Given \:-4 \: is \: a \: one \: of \: the \:zero \:of$

$p(x)$

$p(-4) = 0$

$\implies (k-1)(-4)^{2} + k(-4) + 1 =0$

$\implies (k-1)\times 16 - 4k + 1 = 0$

$\implies 16k - 16 - 4k + 1 = 0$

$\implies 12k - 15 = 0$

$\implies 12k = 15$

$\implies k = \frac{15}{12}$

$\implies k = \frac{5}{4}$

Therefore.,

$\red{ Value \:of \: k } \green { = \frac{5}{4} }$

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Answer 2

Answer:

k = 5/4

Step-by-step explanation:

=> p(x) = (k-1)x²+kx+1

p(x)=(p-4)=0

=> (k-1) (-4)² + k(-4) + 1 = 0

=> (k-1) × 16 - 4k + 1 = 0

=> 16k - 16 - 4k + 1 = 0

=> 12k - 15 = 0

=> 12k = 15

=> k = 15/12

=> k = 5/4

Therefore, value of k = 5/4

hope it helps ! ♥️