Question

Determine the number and type of solutions of the quadratic

equation.
4x2 – 3x + 1 = 0
O one real solution
O two imaginary solutions
O one imaginary solution
O two real solutions

Answer 1

4x2 – 3x + 1 =

• 0 one real solution
• two imaginary solutions
• one imaginary solution
• two real solutions

→ two imaginary solutions

Answer 2

Answer: Choice B

Two imaginary solutions

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Step-by-step explanation:

The given equation is

4x^2 - 3x + 1 = 0

Compare this to the general quadratic

ax^2 + bx + c = 0

to find that a = 4, b = -3, c = 1.

Plug those values into the discriminant formula below.

d = b^2 - 4ac

d = (-3)^2 - 4*4*1

d = 9 - 16

d = -7

Then recall that...

• If d > 0, then we have two real solutions.
• If d = 0, then we have only one real solution.
• If d < 0, then we have two imaginary solutions.

We see that d = -7 fits with the third option highlighted above.