Question

if a² + b² = 117 and ab = 54,then a+b/a-b is

Answer 1

a²+b² = 117
ab = 54
as a²+b² = (a+b)²-2ab
therefore
117 = (a+b)²-2*54
117 = (a+b)²-108
117+108 = (a+b)²
225 = (a+b)²
15 = (a+b) ___ (i)
as a²+b² = (a-b)²+2ab
therefore
117 = (a-b)²+2*54
117 = (a-b)²+108
117-108 = (a-b)²
9 = (a-b)²
3 = (a-b) ___ (ii)
from (i) & (ii)
(a+b)/(a-b) = 15/3
(a+b)/(a-b) = 5

Answer 2

Answer:

The value of expression is $\frac{a+b}{a-b}=5$        

Step-by-step explanation:

Given : If $a^2+b^2=117$ and $ab=54$

To find : The value of $\frac{a+b}{a-b}$

Solution :

Let $x=\frac{a+b}{a-b}$

Squaring both side,

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

$x^2=\frac{(a+b)62}{(a-b)^2}$

$x^2=\frac{a^2+b^2+2ab}{a^2+b^2-2ab}$

Substitute the value given,

$x^2=\frac{117+2(54)}{117-2(54)}$

$x^2=\frac{117+108}{117-108}$

$x^2=\frac{225}{9}$

$x^2=25$

$x=\sqrt{25}$

$x=5$

Therefore, The value of expression is $\frac{a+b}{a-b}=5$