If x - y - 4 and xy - 12 where x-y then the value of x2-y24 will be

Given, x y=4 On squaring both side we get, (x y)^2=4^2 On applying identity we get, ⇒ x^2 2xy+y^2=16 ⇒ x^2 2× 12+y^2=16 nbsp; nbsp; nbsp; [Given xy=1 ...

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Byju's Answer

Standard VIII

Mathematics

Algebraic Identity: (x - y)^2

If x - y = 4 ...

Question

If $x - y = 4$ and $x y = 12$ where $x > y$ then the value of $\frac{(x^{2} + y^{2})}{4}$ will be

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Solution

Given,
$x - y = 4$

On squaring both side we get,
$(x - y)^{2} = 4^{2}$

On applying identity we get,
$\Rightarrow x^{2} - 2 x y + y^{2} = 16$

$\Rightarrow x^{2} - 2 \times 12 + y^{2} = 16$
[Given $x y = 12$ ]

$\Rightarrow x^{2} + y^{2} = 16 + 24$

$\Rightarrow x^{2} + y^{2} = 40$

On dividing both side by $4$ we get,
$\Rightarrow \frac{x^{2} + y^{2}}{4} = 10$

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If x−y=4 and xy=12 where x>y then the value of (x2+y2)4 will be

Given, x−y=4 On squaring both side we get, (x−y)2=42 On applying identity we get, ⇒x2−2xy+y2=16 ⇒x2−2×12+y2=16 [Given xy=12] ⇒x2+y2=16+24 ⇒x2+y2=40 On dividing both side by 4 we get, ⇒x2+y24=10

If $x - y = 4$ and $x y = 12$ where $x > y$ then the value of $\frac{(x^{2} + y^{2})}{4}$ will be