What are the solutions to the equation 100(1+x)^2=121 ?

The solutions to the equation 100(1+x)^2=121 are x= 1/10 ,x=-21/10

Find the zeros of 100(1+x)^2=121

The zeros of 100(1+x)^2=121 are x= 1/10 ,x=-21/10

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100 (1 + x)^{2} = 121

x = \frac{1}{10}, x = - \frac{21}{10}

Solution steps

100 (1 + x)^{2} = 121

Divide both sides by

100

(1 + x)^{2} = \frac{121}{100}

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

1 + x = \frac{121}{100} : x = \frac{1}{10}

Solve

1 + x = - \frac{121}{100} : x = - \frac{21}{10}

The solutions to the quadratic equation are:

x = \frac{1}{10}, x = - \frac{21}{10}

- 6 v^{2} + 2 x - 12 v + vx

- 14 < 2 - 4 n \leq 17

5 x + 3 = x + 11

(6 - 11) (6 + 11)

- 8.5 x + 0.61 = - 2.79

What are the solutions to the equation 100(1+x)^2=121 ?
The solutions to the equation 100(1+x)^2=121 are x= 1/10 ,x=-21/10
Find the zeros of 100(1+x)^2=121
The zeros of 100(1+x)^2=121 are x= 1/10 ,x=-21/10