Question

If,ab=21 and (a+b)^2/(a−b)^2=25/4, then the value of a^2+b^2+3ab is equal to?​

Answer 1

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Answer 2

$a^2+b^2+3ab=121$

Step-by-step explanation:

$ab=21$

$b=\frac{21}{a}$

$\frac{(a+b)^2}{(a-b)^2}=\frac{25}{4}$

$(\frac{a+b}{a-b})^2=(\frac{5}{2})^2$

$\frac{a+b}{a-b}=\pm \frac{5}{2}$

$\frac{a+b}{a-b}=\frac{5}{2}$

$2a+2b=5a-5b$

$2b+5b=5a-2a$

$7b=3a$

$7\times \frac{21}{a}=3a$

$147=3a^2$

$a^2=\frac{147}{3}=49$

$a=\sqrt49}=\pm 7$

$b=\frac{21}{7}=3$

$b=\frac{21}{-7}=-3$

$\frac{a+b}{a-b}=-\frac{5}{2}$

$2a+2b=-5a+5b$

$2a+5a=5b-2b$

$7a=3b$

$7a=3\times \frac{21}{a}$

$a^2=\frac{3\times 21}{7}=9$

$a=\pm 3$

$b=\frac{21}{3}=7$

$b=\frac{21}{-3}=-7$

a=7, b=3

$a^2+b^2+3ab=7^2+3^2+3(7)(3)=121$

a=-7 and b=-3

$a^2+b^2+3ab=(-7)^2+(-3)^2+3(-7)(-3)=121$

a=3 and b=7

$a^2+b^2+3ab=3^2+7^2+3(3)(7)=121$

a=-3 and b=-7

$a^2+b^2+3ab=(-3)^2+(-7)^2+3(-3)(-7)=121$

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