Super simultaneous equations #4

From the first equation, $x^2=x+20$

From the second, $x^2=-5x-4$

So, $\begin{aligned} x+20&=-5x-4 \\ 6x &=-24 \\ x&=\boxed{-4}\end{aligned}$

$(x+2\sqrt{5})(x-2\sqrt{5}) = x = x^2-20 \Rightarrow x^2-x-20 = 0 = (x-5)(x+4) \Rightarrow x = -4 \text{ or } 5$

$x(x+5) = -4 \Rightarrow x^2+5x+4 = 0 = (x+4)(x+1) \Rightarrow x = -4 \text{ or } -1$

The only $x$ value which solves both equations is $x=-4$

Let's solve the first equation first: $\begin{aligned} \left(x+2\sqrt{5}\right)\left(x-2\sqrt{5}\right) &= x \\ \xRightarrow{\text{expand brackets}} x^2-20 &= x \\ \xRightarrow{\text{subtract }x} x^2 - x - 20 = 0 \\ \xRightarrow{\text{factorise}} (x+4)(x-5)=0 \end{aligned}$

The 2 solutions to this equation are $-4$ and $5$ . However, we need to see which one fits the second equation:

• Plugging in $5$ into $x(x+5)$ gives us $50$ , not $-4$ .
• Plugging in $-4$ into $x(x+5)$ gives us $-4$ which is correct!

Therefore, the answer is $\green{\boxed{-4}}$