What are the solutions to the equation 4r(r-1)-0.5r=0 ?

The solutions to the equation 4r(r-1)-0.5r=0 are r=0,r=1.125

Find the zeros of 4r(r-1)-0.5r=0

The zeros of 4r(r-1)-0.5r=0 are r=0,r=1.125

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4 r (r - 1) - 0.5 r = 0

r = 0, r = 1.125

Solution steps

4 r (r - 1) - 0.5 r = 0

Factor

4 r (r - 1) - 0.5 r : r (4 r - 4.5)

r (4 r - 4.5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

r = 0 or 4 r - 4.5 = 0

Solve

4 r - 4.5 = 0 : r = 1.125

The solutions to the quadratic equation are:

r = 0, r = 1.125

∣ - 1 - 2 x ∣ > 9

31 j - 40 = 487

- \frac{1}{2} \cdot \frac{3}{5}

\frac{3}{10} + \frac{1}{2}

\frac{x ^{3} + 64}{x ^{2} + 16}

What are the solutions to the equation 4r(r-1)-0.5r=0 ?
The solutions to the equation 4r(r-1)-0.5r=0 are r=0,r=1.125
Find the zeros of 4r(r-1)-0.5r=0
The zeros of 4r(r-1)-0.5r=0 are r=0,r=1.125