Question

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

Answer 1

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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