plz help me soon

Question

Find all values of h

 for which the quadratic equation has one real solution.

5x^2+9x-h=0

Answer 1

When does the Quadratic Equation have a non-real solution?

x = -b ± √(b² -4ac) / 2a

When that middle section (called the discriminant) within the square root is negative, we cannot evaluate it, we say it has no real solution. Instead, we want to know the values of h where there is a real solution, so we want b²-4ac to not be negative.

But wait! It's more than that, we don't just real solutions, we want a single real solution. We get two solutions because of that pesky ± symbol in the formula, where we evaluate it once with a + and once with a -. Because of this we get two solutions... except when the discriminant (b²-4ac) is equal to zero!

Where is b²-4ac = 0?

Answer 2

The discriminant, D should be equal to 0.

D = b² - 4ac

where b = 9, a = 5 and c = -h

Substitute:

b² - 4ac = 0

(9)² - 4(5)(-h) = 0

81+ 20h = 0 ---> from here solve for h.