Question

If the sum of the square of the zeroes of
the polynomial p(x)=x2+7x+k is 25 ,then k
is equal to​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Required value of k is 12.

Step-by-step explanation:

Given,

$p(x) = {x}^{2} + 7x + k.....(1)$ is a polynomial .

We are comparing polynomial (1) with

$s {x}^{2} + qx + r ....(2)$

and get,

$s = 1 \\ q = 7 \\ r = k$

We know if polynomial (2) has two roots then,

sum of two roots $= - \frac{q}{s}$

and product of the roots $= \frac{r}{s}$

Let a and b be two roots of given polynomial (1).

$a + b = - \frac{ 7}{1} \\ a + b = - 7....(3)$

and

$ab = \frac{k}{1} \\ ab = k$

It is given that sum of the square of the zeroes of the polynomial (1) is 25

So,

${a}^{2} + {b}^{2} = 25 \\ {(a + b)}^{2} - 2ab = 25 \\ {( - 7)}^{2} - 2 \times k = 25 \\ 49 - 2k = 25 \\ - 2k = 25 - 49 \\ - 2k = - 24 \\ k = \frac{ - 24}{ - 2} \\ k = 12$

So, required value of k is 12.

Here applied formula,

${a}^{2} + {b}^{2} = {(a + b)}^{2} - 2ab$

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