Question

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

Answer 1

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`