# How do you solve 100x^2 = 121?

Jul 21, 2016

$x = + \frac{11}{10} \mathmr{and} x = - \frac{11}{10}$

#### Explanation:

The general rule for solving a quadratic equation (x^2") is to make it equal to 0.
However this case is the exception because there is no term in $x$.

$100 {x}^{2} = 121 \text{ make "x^2" the subject}$

${x}^{2} = \frac{121}{100} \text{ find the square root}$

$x = \pm \sqrt{\frac{121}{100}}$

$x = + \frac{11}{10} \mathmr{and} x = - \frac{11}{10}$

We can also follow the usual method to get:

$100 {x}^{2} - 121 = 0 \text{ factorise}$

$\left(10 x + 11\right) \left(10 x - 11\right) = 0$

$x = - \frac{11}{1} \mathmr{and} x = \frac{11}{10}$

Jul 21, 2016

$x = 1.1$
$x = - 1.1$

#### Explanation:

$100 {x}^{2} = 121$
OR
${x}^{2} = \frac{121}{100}$
OR
$x = \sqrt{\frac{121}{100}}$
OR
$x = \frac{11}{10}$
$x = 1.1$
OR
$x = - \frac{11}{10}$
$x = - 1.1$