What are the solutions to the equation (0.27+x)(x)=1.27*10^{-36} ?

The solutions to the equation (0.27+x)(x)=1.27*10^{-36} are x=-0.27,x=0

Find the zeros of (0.27+x)(x)=1.27*10^{-36}

The zeros of (0.27+x)(x)=1.27*10^{-36} are x=-0.27,x=0

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(0.27 + x) (x) = 1.27 \cdot 1 0^{- 36}

x = - 0.27, x = 0

Solution steps

(0.27 + x) (x) = 1.27 \cdot 1 0^{- 36}

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

0.27 + x = 0 or x = 0

Solve

0.27 + x = 0 : x = - 0.27

The solutions to the quadratic equation are:

x = - 0.27, x = 0

4 b^{4} + 20 b^{3} + 12 b

0 = 64 + 80 t - 16 t^{2}

5 x + 8 = 28

4 v^{2} - 100 = 0

27 \cdot 5 \cdot 27 \cdot 10 \cdot 10 \cdot 10 \cdot 10

What are the solutions to the equation (0.27+x)(x)=1.27*10^{-36} ?
The solutions to the equation (0.27+x)(x)=1.27*10^{-36} are x=-0.27,x=0
Find the zeros of (0.27+x)(x)=1.27*10^{-36}
The zeros of (0.27+x)(x)=1.27*10^{-36} are x=-0.27,x=0