What are the solutions to the equation (x)(x+1)=1.5x^2 ?

The solutions to the equation (x)(x+1)=1.5x^2 are x=0,x=2

Find the zeros of (x)(x+1)=1.5x^2

The zeros of (x)(x+1)=1.5x^2 are x=0,x=2

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(x) (x + 1) = 1.5 x^{2}

x = 0, x = 2

Solution steps

(x) (x + 1) = 1.5 x^{2}

Move

1.5 x^{2}

to the left side

x (x + 1) - 1.5 x^{2} = 0

Factor

x (x + 1) - 1.5 x^{2} : - x (0.5 x - 1)

- x (0.5 x - 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or 0.5 x - 1 = 0

Solve

0.5 x - 1 = 0 : x = 2

The solutions to the quadratic equation are:

x = 0, x = 2

7 x^{2} - 5 x = 0

(x - 5)^{2} = 16

- 63 = - 7 c

\frac{- 3 x + 1}{3} + \frac{5}{3} = - \frac{5 x}{4}

161

What are the solutions to the equation (x)(x+1)=1.5x^2 ?
The solutions to the equation (x)(x+1)=1.5x^2 are x=0,x=2
Find the zeros of (x)(x+1)=1.5x^2
The zeros of (x)(x+1)=1.5x^2 are x=0,x=2