Question

If a-b= 1, a²+b² = 13 find ab

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Answer 1

ANSWERGiven, a + b = 5 and a 2 +b 2 =13On squaring a + b = 5 both sides, we get(a+b) 2 =(5) 2 a 2 +b 2 +2ab=2513+2ab=25⇒2ab=25−13=12⇒ab=12/2=6∴ab=6

Answer 2

Solution :

$a - b = 1 \\ \\ a = 1 + b \: \: \: \: \: .....(1)$

$now \: substitute \: eq \: 1 \: \: in \: {a}^{2} + {b}^{2} = 13$

${(1 + b)}^{2} + {b}^{2} = 13$

$1 + {b}^{2} + {b}^{2} = 13$

$2 {b}^{2} = 13 - 1$

$2 {b}^{2} = 12$

${b}^{2} = \frac{12}{2}$

${b}^{2} = 6$

$b = \sqrt{6}$

$a = 1 + \sqrt{6} = 1 \sqrt{6}$

$ab = \sqrt{6} \times 1 \sqrt{6} = 3.83$