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Is it true that if the sum of squares of the zeroes of the quadratic polynomial $x^{2} - 8 x + k$ is 40, then $= 12$ ?

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True

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answer is A.

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Detailed Solution

Given quadratic polynomial,

x^{2} - 8 x + k

. Let

α and β

be the zeroes of

x^{2} - 8 x + k

then according to question,

α^{2} + β^{2} = 40

.
We know that, the relationship between zeroes and coefficient of quadratic polynomial is

Sum of Zeroes = - \frac{Coefficient of x}{Coefficient of x^{2}}

and

Product of Zeroes = \frac{Constant Term}{Coefficient of x^{2}}

.
Then, we get,

α + β = - \frac{(- 8)}{1} = \frac{8}{1}

and

αβ = \frac{k}{1}

.
Now,

α + β = 8

. On squaring both sides,

{(α + β)}^{2} = 8^{2}

{

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

}

\Rightarrow α^{2} + β^{2} + 2 αβ = 64

\Rightarrow 40 + 2 k = 64

\Rightarrow 2 k = 64 - 40

\Rightarrow 2 k = 24

\Rightarrow k = 12

Therefore, the given statement is true.
Hence, option (1) is correct.

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