Question

Solve by completing the square
(4-e)^(2)+(6-e)^(2)=20

Answer 1

`\boxed{e=8, e=2}`

Explanation:

1. Expand
`\large \displaystyle \left(4-e\right)^2\:+\left(6-e\right)^2`

Expand each term in the expression using the identity

`(a-b)^2 = a^2 -2ab+b^2`

`(4 - e)^2= 16-8e+e^2\\\\\left(6-e\right)^2 = 36-12e+e^2`

`\large \displaystyle \left(4-e\right)^2\:+\left(6-e\right)^2 = 16-8e+e^2 + 36-12e+e^2`

Combining like terms we get

`(16 + 36) -(8e+12e) +(e^2 + e^2) = 20`

`\longrightarrow 52 - 20e + 2e^2 = 20`

Subtract 52 from both sides and rearrange terms:

`\longrightarrow 2e^2 - 20e = -32 \\\\`

Divide both sides by 2

`(2e^2-20e)/(2)=(-32)/(2)`

`e^2-10e=-16`

To rewrite in the form (x+a)² = b

we must have 2ae = -10e and therefore a = -5

So the expression

`e^2-10e=-16`

can be transformed by

Adding (-5)² to both sides to complete the square:

`e^2-10e+\left(-5\right)^2=-16+\left(-5\right)^2`

`e^2-10e+\left(-5\right)^2=\left(e-5\right)^2`

`-16 + (-5)^2 = -16 + 25 = 9\\\\`\

So we get

`\left(e-5\right)^2=9`

`\longrightarrow e-5 = \pm √(9) \\\\e - 5 = \pm 3\\\\\\e-5 = + 3 \implies e = 5 + 3 = 8\\\\e-5 = -3 \implies e = -3 + 5 = 2\\\\`

So the solution set is

`\boxed{e=8, e=2}`