Answered: "How do you solve 6x2−4=5x?"

High School
Algebra
Algebra I
Quadratic function and equation

Calvin Hess

Answered question

2022-01-20

How do you solve

6 x^{2} - 4 = 5 x

Answer & Explanation

Messiah Haynes

Beginner2022-01-21Added 7 answers

Step 1

6 x^{2} - 4 = 5 x

Gather all terms on the left side.

6 x^{2} - 5 x - 4 = 0

This is a quadratic equation,

a x^{2} + b x + c

, where

a = 6, b = - 5

, and

c = - 4

You can use the quadratic formula to solve this equation.

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

x = \frac{- (- 5) \pm \sqrt{- 5^{2} = 4 \times 6 \times - 4}}{2 \times 6} =

x = \frac{5 \pm \sqrt{25 + 96}}{12} =

x = \frac{5 \pm \sqrt{121}}{12} =

x = \frac{5 \pm 11}{12}

Solve for x.

x = \frac{5 + 11}{12} = \frac{16}{12} = \frac{4}{3}

x = \frac{5 - 11}{12} = - \frac{6}{12} = - \frac{1}{2}

x = - \frac{1}{2}, \frac{4}{3}

Maritza Mccall

Beginner2022-01-22Added 17 answers

Step 1
Solve
1) $y = 6 x^{2} - 5 x - 4 = 0$
I use the new Transforming Method.
Transformed
2) $y^{'} = x^{2} - 5 x - 24$
Roots have opposite signs.
Factor pairs of
$(- 24) \to (- 2, 12) (- 3, 8)$
This sum is $5 = - b$ . Two real roots of (2) are:
$y_{1} = - 3 and y_{2} = 8$
Back to original equation (1), the 2 real roots are:
$x_{1} = y \frac{1}{a} = - \frac{3}{6} = - \frac{1}{2}$ and
$x_{2} = y \frac{2}{a} = \frac{8}{6} = \frac{4}{3}$

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