Question

please give me its solution

Answer 1

the roots of eq.

x^2 +2x - 143 = 0

x^2 + 13x - 11x - 143

x(x + 13) - 11(x + 13)

(x - 11) (x+ 13)

are two root

and sum of sq.

(x-11)^2 + (x+13)^2
x^2 + 121 -22x + x^2 +169 -26x
2x^2 -48x + 290

Answer 2

$\textsf{ \underline{ \underline{First step :}}}$

Finding the roots of the equation –

$\: \: \: \: \: \: \: {\texttt{x}}^{2} + 2\texttt{x} - 143 = 0 \\ \\ \rightarrow \: {\text{x}}^{2} + (13 - 11)\text{x} - 143 = 0 \\ \\ \rightarrow \: {\text{x}}^{2} + 13\text{x} - 11\text{x} - 143 = 0 \\ \\ \rightarrow \: \text{x}(\text{x} + 13) - 11(\text{x} + 13) = 0 \\ \\ \rightarrow \: (\text{x} + 13)(\text{x} - 11)$

So, the roots of the equation [ x² + 2x –143 = 0 ] is [( x + 13 )( x – 11) ]

$\textsf{ \underline{ \underline{Second step :}}}$

The Sum of the squares of ( x + 13 ) and ( x – 11)

$\: \: \: \: \: \: ({\text{x} + 13})^{2} + ( {\text{x} - 11})^{2} \\ \\ \rightarrow \: ( {\text{x}}^{2} + {13}^{2} + 2 \times 13 \times \text{x}) + ({\text{x}}^{2} + {11}^{2} - 2 \times 11 \times \text{x}) \\ \\ \rightarrow \: {\text{x}}^{2} + 169 + 26{\text{x}} \: + {\text{x}}^{2} + 121 - 22{\text{x}} \\ \\ \rightarrow \: 2 {\text{x}}^{2} + 290 + 4{\text{x}}$

$\boxed{2 {\text{x}}^{2} + 4{\text{x}} + 290 }$