Question

Consider the following two quadratic equations.

4x ^ 2 - 12x + 4 = - 4
- x ^ 2 + 4x - 3 = 0
What solution is true for BOTH equations?

Answer 1

1. Move all terms to the left side and set equal to zero. Then set each factor equal to zero. x=2,1

2. Move all terms to the left side and set equal to zero. Then set each factor equal to zero.

x=3,1

Explanation:

Hope it is helpful.....

Answer 2

x = 1

Explanation:

Solving the equations

4x² - 12x + 4 = - 4 ( add 4 to both sides )

4x² - 12x + 8 = 0 ( divide terms by 4 )

x² - 3x + 2 = 0 ← in standard form

(x - 1)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 2 = 0 ⇒ x = 2

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- x² + 4x - 3 = 0 ( multiply through by - 1 )

x² - 4x + 3 = 0 ← in standard form

(x - 1)(x - 3) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x - 3 = 0 ⇒ x = 3

The solution that is true for both equations is x = 1