# Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$

Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$ So we have $$\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$$ The inequality gives $$x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$$ I am not sure how to deal with the equation. Thank you in advance!

• Simlify the equation into the form $\sqrt{5(3x-1)^2-4}=8-(3x-1)^2$ and set a substitution $y=(3x-1)^2$. Dec 26, 2020 at 22:06

Denote $$t:=\sqrt{45x^2-30x+1}$$. Then we observe that $$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$ As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$ Since $$t\geq 0$$, it follows that $$t=4$$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$ Hence $$(3x+1)(x-1)=0$$. Either $$x=1$$ or $$x=-1/3$$.

• Thank you for the response! I don't see how $t=-\dfrac{t^2-36}{5}.$ Dec 26, 2020 at 23:04
• This follows from my first observation $t^2=-5(7+6x-9x^2)+36$, which entails $7+6x-9x^2=-\dfrac{t^2-36}{5}$. I just substitute it and $t=\sqrt{45x^2-30x+1}$ back to the original formula. Dec 26, 2020 at 23:16

Let $$u=9x^2-6x$$ and rewrite the equation as

$$\sqrt{5u+1}=7-u$$

Squaring both sides gives $$5u+1=49-14u+u^2$$, or

$$u^2-19u+48=(u-16)(u-3)=0$$

We see that $$u=16$$ is not a solution, since $$\sqrt{81}\not=-9$$. This leaves us with $$u=3$$, which is a valid solution, since $$\sqrt{16}=4$$, and from this we have

$$9x^2-6x=3\implies3x^2-2x-1=(x-1)(3x+1)=0$$

which gives $$x=1$$ and $$x=-1/3$$ as the complete solution set.

Remark: What makes this work so nicely is that $$45:30=9:6$$. If the coefficients of $$x^2$$ and $$x$$ on the two sides hadn't been in proportion, the solution would have been much more involved.

$$45x^2-30x+1=(-9x^2+6x+7)^2$$ $$-5(-9x^2+6x+7)+36=(-9x^2+6x+7)^2$$ Substitute $$u=-9x^2+6x+7$$ and solve the quadratic equation.

From $$45x^2-30x+1=(7+6x-9x^2)^2$$, you have that $$81x^{4}-108x^{3}-135x^{2}+114x+48 = 0$$

This factors as $$3(x-1)(3x+1)(9x^2-6x-16) = 0$$

So the roots of that polynomial are $$x = 1, -\frac{1}{3}, \frac{1\pm\sqrt{17}}{3}$$. Out of these, the only ones in the range $$\left[\frac{1-2\sqrt{2}}{3} ,\frac{1+2\sqrt{2}}{3} \right]$$ are $$x = 1$$ and $$x = -\frac{1}{3}$$.