# How to Find x given y of Quadratic Functions

## Question 1

Find any values of $x$ for $y = x^2+x$ if $y=2$.

\begin{align} x^2+x &= 2 \\ x^2+x-2 &= 0 \\ (x-1)(x+2) &= 0 \\ \therefore x &=1 \text{ or } x=-2 \end{align}

## Question 2

Find any values of $x$ for $y = x^2-6x+10$ if $y=1$.

\begin{align} x^2-6x+10 &= 1 \\ x^2-6x+9 &= 0 \\ (x-3)^2 &= 0 \\ \therefore x=3 \end{align}

## Question 3

Find any values of $x$ for $y = 3x^2$ if $y=12$.

\begin{align} 3x^2 &= 12 \\ x^2 &= 4 \\ x &= \pm \sqrt{4} \\ \therefore x &= \pm 2 \end{align}

## Question 4

Find any values of $x$ for $y = x^2+2x$ if $y=1$.

\begin{align} \displaystyle x^2+2x &= 1 \\ x^2+2x-1 &= 0 \\ x &= \frac{-2 \pm \sqrt{2^2-4 \times 1 \times -1}}{2 \times 1} \\ &= \frac{-2 \pm \sqrt{8}}{2} \\ &= \frac{-2 \pm 2\sqrt{2}}{2} \\ \therefore x &= -1 \pm \sqrt{2} \end{align}

## Question 5

Find any values of $x$ for $y = x^2+2x+3$ if $y=1$.

\begin{align} \displaystyle x^2+2x+3 &= 1 \\ x^2+2x+2 &= 0 \\ x &= \frac{-2 \pm \sqrt{2^2-4 \times 1 \times 2}}{2 \times 1} \\ &= \frac{-2 \pm \sqrt{-4}}{2} \\ \therefore &\text{No real values for } x \end{align}