Question

28.
If the sum of the squares of zeros of the polynomial
11
5x2 + 3x + k is -11/25, find value of k​

Answer 1

$\alpha + \beta = \frac{ - b}{a} = \frac{ - 3}{5}$

$\alpha \beta = \frac{c}{a} = \frac{k}{5}$

${ \alpha }^{2} + { \beta }^{2} = \frac{ - 11}{25}$

${ (\alpha + \beta) }^{2} = { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2}$

${( \frac{ - 3}{5}) }^{2} = \frac{ - 11}{25} + 2( \frac{k}{5} )$

$\frac{9}{25} = \frac{ - 11}{25} + \frac{2k}{5}$

$\frac{9}{25} = \frac{ - 11 + 10k}{25}$

$9 + 11 = 10k$

$20 = 10k$

$k = \frac{20}{10}$

$k = 10$

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