What are the solutions to the equation (2t-1)^2(t+4)=0 ?

The solutions to the equation (2t-1)^2(t+4)=0 are t= 1/2 ,t=-4

Find the zeros of (2t-1)^2(t+4)=0

The zeros of (2t-1)^2(t+4)=0 are t= 1/2 ,t=-4

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

t = \frac{1}{2}, t = - 4

Decimal Notation

t = 0.5, t = - 4

Solution steps

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 t - 1 = 0 or t + 4 = 0

Solve

2 t - 1 = 0 : t = \frac{1}{2}

Solve

t + 4 = 0 : t = - 4

The solutions are

t = \frac{1}{2}, t = - 4

y, x^{2} = y^{3}

2 x^{3} + 32 = 8 x^{2} + 8 x

(x) = 3 x^{3} - 12 x^{2} + 8 x + 2

- x^{2} (x - 2) (2 - x) = 0

x^{3} - 5 = - 7

What are the solutions to the equation (2t-1)^2(t+4)=0 ?
The solutions to the equation (2t-1)^2(t+4)=0 are t= 1/2 ,t=-4
Find the zeros of (2t-1)^2(t+4)=0
The zeros of (2t-1)^2(t+4)=0 are t= 1/2 ,t=-4