What are the solutions to the equation (x^2)/8-4= x/2 ?

The solutions to the equation (x^2)/8-4= x/2 are x=8,x=-4

Find the zeros of (x^2)/8-4= x/2

The zeros of (x^2)/8-4= x/2 are x=8,x=-4

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\frac{x ^{2}}{8} - 4 = \frac{x}{2}

x = 8, x = - 4

Solution steps

\frac{x ^{2}}{8} - 4 = \frac{x}{2}

Find Least Common Multiplier of

8, 2 : 8

Multiply by LCM=

8

\frac{x ^{2}}{8} \cdot 8 - 4 \cdot 8 = \frac{x}{2} \cdot 8

Simplify

x^{2} - 32 = 4 x

Move

4 x

to the left side

x^{2} - 32 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 4 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot (- 32) = 12

x_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 12}{2 \cdot 1}

x = \frac{- ( - 4 ) + 12}{2 \cdot 1} : 8

x = \frac{- ( - 4 ) - 12}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

x = 8, x = - 4

- 1

(x - 2)^{2} > - 10

\frac{K - 1}{3} \geq 1

2 - 4 (x + 1) \geq 3 (x - 1) - 13

- 2 + (- 2)

What are the solutions to the equation (x^2)/8-4= x/2 ?
The solutions to the equation (x^2)/8-4= x/2 are x=8,x=-4
Find the zeros of (x^2)/8-4= x/2
The zeros of (x^2)/8-4= x/2 are x=8,x=-4