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$3x-4y = 10$ and $xy = -1$, find the value of $(9x^2 + 16y^2)$.
Given:
$3x -4y = 10$ and $xy = -1$
To do:
We have to find the value of $9x^2 + 16y^2$.
Solution:
We know that,
$(a+b)^2=a^2+b^2+2ab$
$(a-b)^2=a^2+b^2-2ab$
$(a+b)(a-b)=a^2-b^2$
Therefore,
$3x - 4y = 10$
Squaring both sides, we get,
$(3x - 4y)^2 = (10)^2$
$(3x)^2 + (4y)^2 - 2 \times 3x \times 4y = 100$
$9x^2 + 16y^2 - 24xy=100$
$9x^2 + 16y^2 =100+24(-1)$
$9x^2 + 16y^2 = 100 - 24$
$9x^2 + 16y^2 = 76$
The value of $9x^2+16y^2$ is $76$.
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