Question

if one of the two zeroes of quadratic polynomial (k+2)2x^2+kx+4is -4 then find the value of k
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Answer 1

If (-4) is a zero then,

P(-4) = 0

=> (k+2)2(16) - 4k + 4 = 0

=> 32k + 64 - 4k - 4 = 0

=> 28k = (-60)

=> k = (-15)/7 or (-2.5)

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Answer 2

Answer:

The value of k in the given polynomial is $-\dfrac{17}{7}$.

Explanation:

Given a quadratic polynomial $(k+2)2x^2+kx+4$

One of the zeroes of the polynomial is -4.

The zero of a polynomial refers to those values of the variable for which the polynomial equals to zero.

So, the variable in the given polynomial is x and when we substitute $x=-4$ in the polynomial, the result should equal to zero i.e., $P(-4) = 0$.

Given the polynomial, $(k+2)2x^2+kx+4$

substituting $x=-4$,

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

$\implies (2k+4) 16 -4k + 4 = 0$

$\implies 32k+64 -4k + 4 = 0$

  $\implies 28k+68 = 0$  

$\implies 28k=-68 \implies k=- \frac{68}{28}$  

$\implies k=- \frac{68}{28}=- \frac{17}{7}$

Therefore, the value of k in the given polynomial is $-\dfrac{17}{7}$.