Question

can any one solve this all question right now i am gettin bored

Answer 1

Hope u like my process
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16)

Given,
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$p(x) = k {x}^{2} + 4x + 4 \\ \\ or. \: \: p(x) = {x}^{2} + \frac{4}{k} x + \frac{4}{k}$
$\alpha \: \: \: and \: \: \: \beta \: \: \: are \: \: \: the \: \: \: two \: \: \: roots$
And

${ \alpha }^{2} + { \beta }^{2} = 24$
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We know,
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For a quadratic equation,

x² - (sum of the roots) + (product of the roots)=0

So,

Sum of the roots = - 4/k

$i.e. \: \: \: \alpha + \beta = - \frac{4}{k}$

Product of the roots = 4/k

$i.e. \: \: \: \alpha \beta = \frac{4}{k}$
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Now.,
$= > { \alpha }^{2} + { \beta }^{2} = 24 \\ \\ or. \: {( \alpha + \beta )}^{2} - 2 \alpha \beta = 24 \\ \\ or. \: \: {( - \frac{4}{k} )}^{2} - 2( \frac{4}{k} ) = 24 \\ \\ or. \: \: \frac{16}{ {k}^{2} } = 24 + \frac{8}{k} \\ \\ or. \: \: 8( \frac{2}{ {k}^{2} } ) = 8( \frac{3k + 1)}{k} ) \\ \\ or. \: \: \frac{2}{ {k}^{2} } \times \frac{8k}{8} = 3k + 1 \\ \\ or. \: \: 2 = 3 {k}^{2} + k \\ \\ or. \: \: 3 {k}^{2} + k - 2 = 0 \\ \\ or. \: \: 3 {k}^{2} + 3k - 2k - 2 = 0 \\ \\ or. \: \: 3k(k + 1) - 2(k + 1) = 0 \\ \\ or. \: \: (3k - 2)(k + 1) = 0 \\ \\ either \: \: \\ ........... \\ k = \frac{2}{3} \\ \\ or \\ .... \\ k = - 1$
So

K = -1 , ⅔
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Hope this is ur required answer

Proud to help you

Answer 2

Hey these all were solved !!!!


Yoo !!!!