Question

if one zero of the quadratic polynomial is (k-1), x^2+kx+1 is -4 then the value of k is​

Answer 1

Step-by-step explanation:

$\huge\underline{\underline{\ Question}}$

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If one zero of the quadratic polynomial is (k-1), and x^2+kx+1 = -4 then the value of k is?

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$\huge\underline{\underline{\ Answer}}$

If (k-1) is zero (root) of following equation,

then it will satisfy the equation

${x}^{2} + kx + 1 = - 4 \\ {x}^{2} + kx - 3 = 0 \\ put \: x =( k - 1 )\\ {k}^{2} + 1 - 2k + {k}^{2} - k - 3 = 0 \\ 2 {k}^{2} - 3k - 2 = 0 \\ k = \frac{3 + \: or \: - \sqrt{25} }{2} \\ k = \frac{3 + \: or \: - 5}{2}\\ \huge \boxed{k = 4 \: \: \: or \: \: \: k = -1}$

Answer 2

Correct Question

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

Answer:

$\boxed{\boxed{\mathtt{k = \dfrac{5}{4}}}}$

Step-by-Step Explanation:

• Given

One Zero of Polynomial is -4

$\mathsf{P(X) = (k - 1) {x}^{2} + kx + 1}$

• To find

Value of K.

• Solution

Since it's given that -4 is a Zero of p(X) therefore -4 is a Factor of Polynomial P(X).

Now by Factor Theorem

" If x - a is a Factor of Polynomial P(X) then p(a) = 0 "

*Therefore here -4 is a Factor of the polynomial *

$\mathtt{p(-4) = 0}$

That is,

$\implies \mathsf{P(X) = (k - 1) {x}^{2} + kx + 1 = 0}$

$\longrightarrow \mathsf{P(-4) = (k - 1) {(-4)}^{2} + k(-4) + 1 = 0}$

$\mathsf{\longrightarrow P(-4) = (k - 1)(16) - 4k + 1 = 0}$

$\mathsf{ \longrightarrow = 16k - 16 - 4k + 1 = 0}$

$\mathsf{ \longrightarrow 12k -15 = 0}$

$\longrightarrow \mathsf{ 12k = 15 }$

$\longrightarrow \mathsf{ k = \dfrac{15}{12} }$

$\longrightarrow \mathsf{ k = \dfrac{5}{4}}$

Thus we have required answer =

$\underline{\boxed{\mathtt{k = \dfrac{5}{4}}}}$