h^2+5h-48=0 solve for h

Question

I need the above problem solved for h and h+5

Answer 1

Do you know the quadratic formula? If not i recommend memorizing this.

 

If Ah²+Bh+C = 0 then h = -(B/2A)+(B²-4AC)^1/2/(2A)

In your example A = 1, B = 5 and C = -48. Plug into formula above and you should get 2 answers.

 

If you don't remember the quadratic formula you can always derive it by completing the square.

Ah²+Bh+C = h²+(B/A)h+(C/A) = 0

Complete the square

 

[h+(B/A)/2]²+(C/A)-[(B/A)/2]² = [h+B/(2A)]²+(C/A)-B²/(4A²) = 0

 

Solve for h and you see

[h+B/(2A)]² = B²/(4A²)-C/A = (B²-4AC)/(4A²)

Take square root

|h+B/(2A)| = (B²-4AC)^1/2/(2A)

The absolute value we can remove knowing |h+B/(2A)| = +(h+B/(2A))

Thus

h = -(B/2A)+(B²-4AC)^1/2/(2A)

 

Best - David