Math, asked by ujjwalsrivastava95, 4 months ago

Let x and y be real numbers satisfying the inequality 5x² + y² - 4xy + 24 ≤ 10x - 1. Find the value of x² + y² - 29.​

Answers

Answered by pulakmath007
2

SOLUTION

GIVEN

Let x and y be real numbers satisfying the inequality

5x² + y² - 4xy + 24 ≤ 10x - 1.

TO DETERMINE

The value of x² + y² - 29.

EVALUATION

Here the given inequality is

 \sf 5 {x}^{2}  +  {y}^{2}  - 4xy + 24 \leqslant 10x - 1

 \sf  \implies \: 5 {x}^{2}  +  {y}^{2}  - 4xy + 25 - 10x \leqslant 0

 \sf  \implies \: 4 {x}^{2}    - 4xy  +  {y}^{2} +  {x}^{2}  - 10x + 25 \leqslant 0

 \sf  \implies \:  {(2x)}^{2}    - 2.2x.y  +  {y}^{2}+  {x}^{2}  - 2.x.5 +  {5}^{2}  \leqslant 0

 \sf  \implies \:  {(2x - y)}^{2} +  {(x - 5)}^{2}   \leqslant 0

Since sum of squares of two real numbers can not be negative

So we have

 \sf   \:  {(2x - y)}^{2} +  {(x - 5)}^{2}   =  0

Above gives

 \sf   \:  {(2x - y)}^{}  = 0 \:  \: and \:  \:   {(x - 5)}^{}   =  0

 \sf  \implies \: y = 2x \:  \: and \:  \:   x = 5

 \sf  \implies \: y = 2 \times 5

 \sf  \implies \: y = 10

Thus we have x = 5 , y = 10

Now

 \sf  {x}^{2}  +  {y}^{2}  - 29

 \sf  =  {5}^{2}  +  {10}^{2}  - 29

 \sf  = 25 + 100  - 29

 \sf  = 96

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