How to approach solving for $x$? $(1+x)^2 = 1.21$

Question

I'm trying to solve for $x$ in the following equation, but don't know how to solve it without using a graphing calculator.

$$(1+x)^2 = 1.21$$

Is there a rule I'm supposed to follow?

Answer 1

You can solve this as a general quadratic equation, but the special form makes it easier. In general (for $a \ge 0$): $$x^2 = a \Leftrightarrow x = \pm \sqrt{a}$$

In your case, since $1.1^2 = 1.21$, proceed as follows: $$\left(1+x\right)^2 = 1.21 \Leftrightarrow 1+x = \pm 1.1 \Leftrightarrow x = \ldots$$

Answer 2

Notice, you can also follow this method $$(1+x)^2=1.21$$ $$(1+x)^2=(1.1)^2$$ $$|1+x|=1.1$$ $$1+x=\pm1.1$$ $$x=-1\pm1.1$$ $$x=\color{red}{0.1},\ \color{red}{-2.1} $$

Answer 3

Notice:

• $$x^2=y\Longleftrightarrow x=\pm\sqrt{y}$$
• $$y+x=z\Longleftrightarrow y=z-x$$

So:

$$(a+b)^2=c\Longleftrightarrow$$ $$a+b=\pm\sqrt{c}\Longleftrightarrow$$ $$a=\pm\sqrt{c}-b$$

With your problem we get:

$$(1+x)^2=1.21\Longleftrightarrow$$ $$1+x=\pm\sqrt{1.21}\Longleftrightarrow$$ $$x=\pm\sqrt{1.21}-1$$