What are the solutions to the equation 1/4 =t(1-t) ?

The solutions to the equation 1/4 =t(1-t) are t= 1/2

Find the zeros of 1/4 =t(1-t)

The zeros of 1/4 =t(1-t) are t= 1/2

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\frac{1}{4} = t (1 - t)

t = \frac{1}{2}

Decimal Notation

t = 0.5

Solution steps

\frac{1}{4} = t (1 - t)

Multiply both sides by

4

1 = 4 t (1 - t)

Expand

4 t (1 - t) : 4 t - 4 t^{2}

1 = 4 t - 4 t^{2}

Switch sides

4 t - 4 t^{2} = 1

Move

1

to the left side

4 t - 4 t^{2} - 1 = 0

Write in the standard form

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

- 4 t^{2} + 4 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

4^{2} - 4 (- 4) (- 1) = 0

t_{1, 2} = \frac{- 4 \pm 0}{2 ( - 4 )}

t = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

t = \frac{1}{2}

The solution to the quadratic equation is:

t = \frac{1}{2}

6 x^{2} + 12 = 0

- x^{2} + 9 > 0

(x - 8)^{\frac{7}{2}} - 64 (x - 8)^{\frac{3}{2}}

3 z - 8 + 6 z - 12 = z - 10 + 9 z - 13

45 = \frac{45}{360} \cdot 2 (6)

What are the solutions to the equation 1/4 =t(1-t) ?
The solutions to the equation 1/4 =t(1-t) are t= 1/2
Find the zeros of 1/4 =t(1-t)
The zeros of 1/4 =t(1-t) are t= 1/2