Question

A quadratic equation is shown below

16x^2-8x+1=0

Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work

Part B: Solve 4x^2-8x-45=0 using an appropriate method

Answer 1

Answer:

see explanation

Step-by-step explanation:

A

Given

16x² - 8x + 1 = 0 ← in standard form

with a = 16, b = - 8 and c = 1

Consider the discriminant

Δ = b² - 4ac = (- 8)² - (4 × 16 × 1) = 64 - 64 = 0

This indicates that the equation has 1 real root

B

Using the quadratic formula

x = ( - (- 8) ± $\sqrt{b^2-4ac}$) / (2 × 16)

  = ( 8 ± 0 ) / 32

Hence x = $\frac{8}{32}$ = $\frac{1}{4}$ ← 1 real root