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

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

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

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

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

x = 8, x = - 4

Solution steps

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

Find Least Common Multiplier of

8, 2 : 8

Multiply by LCM=

8

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

Simplify

x^{2} - 4 x = 32

Move

32

to the left side

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

v^{2} + 10 v + 21 = 0

(\frac{1}{2} x + 7)^{2} + 7 (\frac{1}{2} x + 7) + 12 = 0

3 + in (x^{2} + 1) = 2 in (x)

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

(x + 2) (x + 1) - (x - 1) = 6

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