What are the solutions to the equation (x^2)/4-(5x)/2+4=0 ?

The solutions to the equation (x^2)/4-(5x)/2+4=0 are x=8,x=2

Find the zeros of (x^2)/4-(5x)/2+4=0

The zeros of (x^2)/4-(5x)/2+4=0 are x=8,x=2

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\frac{x ^{2}}{4} - \frac{5 x}{2} + 4 = 0

x = 8, x = 2

Solution steps

\frac{x ^{2}}{4} - \frac{5 x}{2} + 4 = 0

Find Least Common Multiplier of

4, 2 : 4

Multiply by LCM=

4

\frac{x ^{2}}{4} \cdot 4 - \frac{5 x}{2} \cdot 4 + 4 \cdot 4 = 0 \cdot 4

Simplify

x^{2} - 10 x + 16 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 16 = 6

x_{1, 2} = \frac{- ( - 10 ) \pm 6}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 6}{2 \cdot 1}, x_{2} = \frac{- ( - 10 ) - 6}{2 \cdot 1}

x = \frac{- ( - 10 ) + 6}{2 \cdot 1} : 8

x = \frac{- ( - 10 ) - 6}{2 \cdot 1} : 2

The solutions to the quadratic equation are:

x = 8, x = 2

35 k^{2} - 22 k + 7 = 4

\frac{x + 2}{2} + 2 = \frac{x + 3}{5}

10 = - 4.9 t^{2} + 18.01 t

2 w^{4} - 7 w^{3} - 15 w^{2}

\frac{- 3}{5} + \frac{- 1}{15} + \frac{8}{25}

What are the solutions to the equation (x^2)/4-(5x)/2+4=0 ?
The solutions to the equation (x^2)/4-(5x)/2+4=0 are x=8,x=2
Find the zeros of (x^2)/4-(5x)/2+4=0
The zeros of (x^2)/4-(5x)/2+4=0 are x=8,x=2