Answer:
I)
[tex]2a + 2b = -10\text{ or } 2a + 2b = 10[/tex]
II)
[tex]2a - 2b = -2 \text{ or } 2a-2b= 2[/tex]
Step-by-step explanation:
We are given that:
[tex]\displaytstyle a^2 + b^2 = 13 \text{ and } ab= 6[/tex]
I)
Recall the perfect square trinomial pattern:
[tex]\displaystyle (a+b)^2 = a^2 + 2ab + b^2[/tex]
Rewrite:
[tex]\displaystyle (a+b)^2 = (a^2 + b^2) + (2ab)[/tex]
Substitute:
[tex]\displaystyle (a+b)^2 = (13) + (12)[/tex]
Evaluate:
[tex]\displaystyle (a + b)^2 = 25[/tex]
Take the square root of both sides:
[tex]\displaystyle a + b = \pm\sqrt{25} = \pm5[/tex]
Hence:
[tex]\displaystyle 2a + 2b = \pm 10[/tex]
Therefore:
[tex]\displaystyle 2a + 2b = -10 \text{ or } 2a + 2b = 10[/tex]
II)
Likewise:
[tex]\displaystyle (a-b)^2 = a^2 - 2ab + b^2[/tex]
Substitute:
[tex]\displaystyle (a-b)^2 = (13) -(12)[/tex]
Solve:
[tex]\displaystyle \begin{aligned} (a-b)^2 &= 1 \\ a-b &= \pm \sqrt{1} \\ a-b &= \pm 1\\ 2a - 2b &= \pm 2\end{aligned}[/tex]
In conclusion:
[tex]2a - 2b = -2 \text{ or } 2a-2b= 2[/tex]
