If a^2+b^2=13 and ab =6 find the value of (i)(2a+2b) (ii)(2a-2b)

Question

asked Jan 20, 2022 129k views

1 Answer

Alex Kucherenko · Answer 1 · 2022-01-26T11:56:54+0000

Answer:

I)

$2a + 2b = -10\text{ or } 2a + 2b = 10$

II)

$2a - 2b = -2 \text{ or } 2a-2b= 2$

Explanation:

We are given that:

$\displaytstyle a^2 + b^2 = 13 \text{ and } ab= 6$

I)

Recall the perfect square trinomial pattern:

$\displaystyle (a+b)^2 = a^2 + 2ab + b^2$

Rewrite:

$\displaystyle (a+b)^2 = (a^2 + b^2) + (2ab)$

Substitute:

$\displaystyle (a+b)^2 = (13) + (12)$

Evaluate:

$\displaystyle (a + b)^2 = 25$

Take the square root of both sides:

$\displaystyle a + b = \pm√(25) = \pm5$

Hence:

$\displaystyle 2a + 2b = \pm 10$

Therefore:

$\displaystyle 2a + 2b = -10 \text{ or } 2a + 2b = 10$

II)

Likewise:

$\displaystyle (a-b)^2 = a^2 - 2ab + b^2$

Substitute:

$\displaystyle (a-b)^2 = (13) -(12)$

Solve:

$\displaystyle \begin{aligned} (a-b)^2 &= 1 \\ a-b &= \pm √(1) \\ a-b &= \pm 1\\ 2a - 2b &= \pm 2\end{aligned}$

In conclusion:

$2a - 2b = -2 \text{ or } 2a-2b= 2$