Solve the equation $\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}$

Question

Solve the equation with $x\in R$$$\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}$$

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My try: $\Leftrightarrow \sqrt{5x^2+4x}-21-\left(\sqrt{x^2-3x-18}-6\right)-\left(5\sqrt{x}-15\right)=0$

$\Leftrightarrow \frac{5x^2+4x-441}{\sqrt{5x^2+4}+21}-\frac{x^2-3x-18-36}{\sqrt{x^2-3x-18}+6}-\frac{25x-225}{5\sqrt{x}+15}=0$

$\Leftrightarrow \frac{\left(x-9\right)\left(5x+49\right)}{\sqrt{5x^2+4}+21}-\frac{\left(x-9\right)\left(x+6\right)}{\sqrt{x^2-3x-18}+6}-\frac{25\left(x-9\right)}{5\sqrt{x}+15}=0$

$\Leftrightarrow \left(x-9\right)\left(\frac{5x+49}{\sqrt{5x^2+4}+21}-\frac{x+6}{\sqrt{x^2-3x-18}+6}-\frac{25}{5\sqrt{x}+15}\right)=0$

$\Rightarrow x=9$ and $\left(\frac{5x+49}{\sqrt{5x^2+4}+21}-\frac{x+6}{\sqrt{x^2-3x-18}+6}-\frac{25}{5\sqrt{x}+15}\right)=0$ i can't solve it. Help me

Answer 1

after squaring and rearranging we get $$3x^2-12x-9=\sqrt{5x^2+4x}\sqrt{x^2-3x-18}$$ squaring again and factorizing we get $$(x-9) (4 x+3) \left(x^2-7 x-3\right)=0$$ Can you finish?

Answer 2

I don't think your approach will allow you to easily find the second solution.

Alternatively, square both sides of: $$\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}$$ To get: $$5x^2+4x+x^2-3x-18-2\sqrt{5x^2+4x}\sqrt{x^2-3x-18}=25x$$ Simplifying and rearranging: $$-2\sqrt{5x^2+4x}\sqrt{x^2-3x-18}=-6 x^2 + 24 x + 18$$ Squaring both sides again: $$4\left( 5x^2+4x \right)\left(x^2-3x-18\right)=\left(-6 x^2 + 24 x + 18\right)^2$$ Rearranging all to one side and factoring (you already found the root $x=9$): $$-4 \left(4 x + 3 \right) \left(x - 9 \right) \left(x^2 - 7 x - 3 \right) = 0$$ Note that the squaring may have introduced extraneous solutions, so be careful.

Answer 3

$$\sqrt{5x^2 + 4x} - \sqrt{x^2 - 3x - 18} = 5\sqrt{x}$$ then squaring both sides, we get $$5x^2 + 4 x + (x^2 - 3x - 18) - 2\sqrt{(5x^2 + 4x) (x^2 - 3x - 18)} = 25 x$$ $$6x^2 + x - 18 - 2\sqrt{(5x^2 + 4x) (x^2 - 3x - 18)} = 25 x$$ $$6x^2 - 24 x - 18 = 2\sqrt{(5x^2 + 4x) (x^2 - 3x - 18)}$$ then squaring both sides, we get $$(6x^2 - 24 x - 18)^2 = 4(5x^2 + 4x) (x^2 - 3x - 18)$$ $$36x^4 - 288x^3 + 360x^2 + 864x + 324 = 20x^4 - 44x^3 - 408x^2 - 288x$$ So that $$16x^4 - 244x^3 + 768x^2 + 1152x + 324 = 0$$ Hence $$4(4x+3)(x-9)(x^2-7x-3) = 0$$ So that the possible values of $x$ are $$x = -\frac{3}{4},9,\frac{7+\sqrt{61}}{2}, \frac{7-\sqrt{61}}{2}$$ We have to substitute these values in the main equation to get the acceptable values of $x$ which are $$x = 9 ,\frac{7+\sqrt{61}}{2}$$

Answer 4

$$\sqrt{5x^2 + 4x} - \sqrt{x^2 - 3x - 18} = 5 \sqrt{x}$$

Restrictions

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$\left[ \begin{array}{rcl} 5x^2 + 4x &\ge & 0 \\ x^2 - 3x - 18 &\ge & 0 \\ x &\ge & 0 \end{array} \right] \implies \left[ \begin{array}{rcl} x \le -\dfrac 45 &\text{or} &x \ge 0 \\ x \le -3 &\text{or} &x \ge 6 \\ & x \ge 0 \end{array} \right] \implies x \le -3 \; \text{or} \; x \ge 6$

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Multiply both sides by $\sqrt{5x^2 + 4x} + \sqrt{x^2 - 3x - 18}$.

\begin{align} (5x^2 + 4x) - (x^2 - 3x - 18) &= 5 \sqrt{x}(\sqrt{5x^2 + 4x} + \sqrt{x^2 - 3x - 18}) \\ \hline 5\sqrt{x}(\sqrt{5x^2 + 4x} + \sqrt{x^2 - 3x - 18}) &= 4x^2 + 7x + 18 \\ 5\sqrt x(\sqrt{5x^2 + 4x} - \sqrt{x^2 - 3x - 18}) &= 25x \\ \hline 10\sqrt x \sqrt{5x^2 + 4x} &= 4x^2 + 32x + 18 \\ 5x\sqrt{5x + 4} &= 2x^2 + 16x + 9 \\ 125x^3 +100x^2 &= 4x^4 + 64x^3 + 292x^2 + 288x + 81 \\ 4 x^4 - 61 x^3 + 192 x^2 + 288 x + 81 &= 0 \\ (4 x + 3) (x - 9) (x^2 - 7 x - 3) &= 0 \end{align}

Applying the restrictions, we get

$$x \in \left\{ \dfrac{7 + \sqrt{61}}{2}, 9 \right\}$$