What are the solutions to the equation x+(x/(100))x=56 ?

The solutions to the equation x+(x/(100))x=56 are x=40,x=-140

Find the zeros of x+(x/(100))x=56

The zeros of x+(x/(100))x=56 are x=40,x=-140

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x + (\frac{x}{100}) x = 56

x = 40, x = - 140

Solution steps

x + (\frac{x}{100}) x = 56

Expand

x + (\frac{x}{100}) x : x + \frac{x ^{2}}{100}

x + \frac{x ^{2}}{100} = 56

Move

56

to the left side

x + \frac{x ^{2}}{100} - 56 = 0

Write in the standard form

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

\frac{1 \cdot x ^{2}}{100} + x - 56 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot \frac{1}{100} ( - 56 )}{2 \cdot \frac{1}{100}}

1^{2} - 4 \cdot \frac{1}{100} (- 56) = \frac{9}{5}

x_{1, 2} = \frac{- 1 \pm \frac{9}{5}}{2 \cdot \frac{1}{100}}

Separate the solutions

x_{1} = \frac{- 1 + \frac{9}{5}}{2 \cdot \frac{1}{100}}, x_{2} = \frac{- 1 - \frac{9}{5}}{2 \cdot \frac{1}{100}}

x = \frac{- 1 + \frac{9}{5}}{2 \frac{1}{100}} : 40

x = \frac{- 1 - \frac{9}{5}}{2 \frac{1}{100}} : - 140

The solutions to the quadratic equation are:

x = 40, x = - 140

81 b^{2} - 64

\frac{x ^{- 2}}{3 x ^{- 3} y ^{2}}

- 1 - 2 (2 x + 1) < x - (- 1 - x)

\frac{1}{2} (3)^{2}

x^{2} - 4 x + 4 = 5

What are the solutions to the equation x+(x/(100))x=56 ?
The solutions to the equation x+(x/(100))x=56 are x=40,x=-140
Find the zeros of x+(x/(100))x=56
The zeros of x+(x/(100))x=56 are x=40,x=-140