Question

solve it fasttt.....,.

Answer 1

GIVEN :–

• Quadratic equation x² - kx + (k² + 7k + 15) = 0.

• Roots of quadratic equation $\implies \bf \:\: x_1 \: and \: x_2 \:\:$

TO FIND :–

• Maximum value of $\bf \:\: x_1^{2} +x_2^{2} = ?\:\:$

SOLUTION :–

• We should write this as –

$\\\implies\bf x_1^{2} +x_2^{2} =(x_1 + x_2)^{2} - 2(x_1)(x_2)\: \: \: \: \:\: - - - -eq.(1)$

• We know that –

$\\\implies\bf Sum \: \: of \: \: roots = - \dfrac{Coefficient \: \: of \: \: x}{Coefficient \: \:of \: \: {x}^{2} }\:\:$

$\\\implies\bf x_1 + x_2= - \dfrac{( - k)}{1}\:\:$

$\\\implies\bf x_1 + x_2=k \: \: \: \: \:\: - - - -eq.(2)$

• And –

$\\\implies\bf Product \: \: of \: \: roots = \dfrac{Constent \: \: term}{Coefficient \: \:of \: \: {x}^{2} }\:\:$

$\\\implies\bf (x_1)( x_2)= \dfrac{( {k}^{2} + 7k + 15)}{1}\:\:$

$\\\implies\bf (x_1)( x_2)={k}^{2} + 7k + 15 \: \: \: \: \:\: - - - -eq.(3)$

• Put the values from eq.(3) & eq.(2) in eq.(1)

$\\\implies\bf x_1^{2} +x_2^{2} =(k)^{2} - 2({k}^{2} + 7k + 15)$

• Let –

$\\\implies\bf x_1^{2} +x_2^{2} =P$

• So that –

$\\\implies\bf P =k^{2} - 2{k}^{2} - 14k - 30$

$\\\implies\bf P = - {k}^{2} - 14k - 30$

• Now Differentiate with respect to 'k' –

$\\\implies\bf \dfrac{dP}{dk} = -2k- 14$

• For Maximum/Minimum –

$\\\implies\bf \dfrac{dP}{dk} = 0$

$\\\implies\bf -2k- 14= 0$

$\\\implies\bf 2k + 14= 0$

$\\\implies\bf 2k = - 14$

$\\\implies \large{ \boxed{\bf k = -7}}$

• Now 'P' –

$\\\implies\bf P_{max} = - {(- 7)}^{2} - 14( - 7) - 30$

$\\\implies\bf P_{max} = -49 + 98 - 30$

$\\\implies\bf P_{max} = 98 - 79$

$\\\implies \large{ \boxed{\bf P_{max} =19}}$

▪︎ Hence , Maximum value of $\bf \:x_1^{2} +x_2^{2} \:$ is 19.

Answer 2

Answer:

K = -7

and

P max = 19.

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