Question

How do you solve the equation `4x^2+9x-1=0` by completing the square?

Answer 1

Solution: `x=1/8(-9+sqrt97) or x=1/8(-9-sqrt97)`

Explanation:

`4x^2+9x-1 = 4(x^2+9/4x) -1 = 4(x^2+9/4x+ (9/8)^2)-81/16 -1` or

`4(x+ 9/8)^2= 81/16 +1 = 4(x+ 9/8)^2= 97/16 = (x+ 9/8)^2= 97/64` or

`(x+ 9/8)= +- sqrt(97/64) or (x+ 9/8)= +- sqrt(97)/8` or

`:. x = -9/8 +- sqrt(97)/8 or x =1/8(-9+-sqrt97)`

Solution: `x=1/8(-9+sqrt97) or x=1/8(-9-sqrt97)` [Ans]

Answer 2

`x=-9/8+-sqrt(97)/8`

Explanation:

Step 1. Factor out the 4 from in front of the `x^2`

`4(x^2+9/4 x-1/4)=0`

Step 2. Take the coefficient of the `x` term (i.e., `9//4`), cut it in half, square it, and add it and subtract it inside the parenthesis.

When you add a number and subtract it, that's like adding `0` and so doesn't change the problem.

Middle term: `9/4`

Cut it in half: `9/8` (this number will be used in Step 3)

Square the result: `(9/8)^2`

Next, add and subtract this term

`4(x^2+9/4 x" "color(red)( + (9/8)^2)color(red)(-(9/8)^2)-1/4)=0`

Step 3. Use the left three terms to make a perfect square

`4(color(blue)(x^2+9/4 x + (9/8)^2)-(9/8)^2-1/4)=0`

`4(color(blue)((x+9/8)^2)-(9/8)^2-1/4)=0`

Step 4. Simplify the fraction and multiply `4` back through.

`4((x+9/8)^2-(9/8)^2-1/4)=0`

`4((x+9/8)^2-97/64)=0` (*Don't forget signs here like I had!)

`4(x+9/8)^2-97/16=0`

Step 5. Use algebra to solve for `x`

`4(x+9/8)^2=97/16`

`(x+9/8)^2=97/64`

`x+9/8=+-sqrt(97/64)`

`x=-9/8+-sqrt(97)/8`