Question

How to solve `5x^2+20x+13=0` by completing the square?

Answer 1

`x=(-10+-sqrt(35))/5`

Explanation:

`5x^2+20x+13=0`
Multiplying both sides by 5, we get
`25x^2+100x+65=0`
`rArr25x^2+2(5x)(10)+100-35=0`
`rArr(5x)^2+2(5x)(10)+(10)^2-35=0`
`rArr(5x+10)^2-35=0`
`rArr(5x+10)^2=35=(sqrt(35))^2`
`rArr5x+10=+-sqrt35`
`5x=-10+-sqrt35`
`x=(-10+-sqrt35)/5`

Answer 2

`5x^2+20x+13=0`

`1/5(x^2 + 4x +13/5) = 0`

`x^2 + 4x +13/5=0`

Now, add and subtract, half of the x-term squared, i.e. `(4/2)^2=2^2 = 4`

`=>[x^2 + 4x +4] +[13/5-4]=0`

`=>[x^2 + 4x +2^2] =[4-13/5]`

`=>(x+2)^2 =20/5-13/5`

`=>(x+2) =+-sqrt(7/5`

`=>x =-2+-sqrt(7/5`