Question

For what values of k will the equation 4x² + 8x + k = 0 have two real solutions?

Answer 1

k < 4

Explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

Then the discriminant Δ = b² - 4ac informs us about the nature of the solutions.

• If b² - 4ac > 0 then 2 real and distinct solutions

• If b² - 4ac = 0 then 1 real and repeated solution

• If b² - 4ac < 0 then no real solutions

4x² + 8x + k = 0 ← is in standard form

with a = 4, b = 8 , c = k , then

b² - 4ac = 8² - (4 × 4 × k) = 64 - 16k

For two real solutions

64 - 16k > 0 ( subtract 64 from both sides )

- 16k > - 64

Divide both sides by - 16, reversing the symbol as a result of dividing by a negative quantity.

k < 4

Then any value less than 4 will ensure the equation has two real solutions