Question 61
If $$a^2 + b^2 = 169, ab = 60, (a > b)$$, then $$(a^2 - b^2)$$ is equal to:
Solution
Given $$a^2 + b^2 = 169, ab = 60, (a > b)$$
$$(a+b)^{2}=a^{2}+b^{2}+2ab$$
$$(a+b)^{2}=169 + 2\times60=289$$
$$(a+b)=17$$
ab=60, so a = $$\frac{60}{b}$$
Now, (a+b)=17
$$\frac{60}{b}$$+b=17
$$b^{2}-17b+60=0$$
$$b^{2}-12b-5b+60=0$$
b(b-12)-5(b-12)=0
b=12 and 5 ...neglect b=12 because a>b given
So,a=12
$$(a^2 - b^2)= (a+b)(a-b)$$=$$17\times7=119$$
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