Question

a+b=3 and a^2+b^2=39find the value of ab

Answer 1

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Answer 2

Hello,

A+b=3 and a^2+b^2=39find the value of ab

a = 3 - b

(3 - b)^2 + b^2 = 39

9 - 6b + b^2 + b^2 = 39

2b^2 - 6b + 9 - 39 = 0

2b^2 - 6b - 30 = 0

$\Delta = 6^2 - 4 \times 2 \times (-30)$

$\Delta = 36 + 240 = 276$

$\sqrt\Delta = \sqrt276$ > 0 so 2 answers

$b_{1} = \dfrac{6 - \sqrt276}{4}$

$b_{2} = \dfrac{6 + \sqrt276}{4}$

a1 = 3 - b1 = b2

a2 = 3 - b2 = b1

ab = $\dfrac{6 - \sqrt276}{4} \times \dfrac{6 + \sqrt276}{4}$

ab = $\dfrac{36 - 276}{16} = \dfrac{-240}{16} = -15$