Question

Solve the following equations.

(i) $\frac{x^2+12x-20}{3x-5}=\frac{x^2+8x+12}{2x+3}$

(ii) $\frac{10x^2+15x+63}{5x^2-25x+12}=\frac{2x+3}{x-5}$

(iii) $\frac{{\left(2x+1\right)}^2+{\left(2x-1\right)}^2}{{\left(2x+1\right)}^2-{\left(2x-1\right)}^2}=\frac{17}{8}$

(iv) $\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$

(v) $\frac{{\left(4x+1\right)}^2+{\left(2x+3\right)}^2}{4x^2+12x+9}=\frac{61}{36}$

(vi) $\frac{{\left(3x-4\right)}^3-{\left(x+1\right)}^3}{{\left(3x-4\right)}^3+{\left(x+1\right)}^3}=\frac{61}{189}$

Answer 1

(i)
$\frac{x^2+12x-20}{3x-5}=\frac{x^2+8x+12}{2x+3}$

Multiplying both sides by $\frac{1}{4}$, we get

$\frac{x^2+12x-20}{12x-20}=\frac{x^2+8x+12}{8x+12}$

Using dividendo, we get

$$\begin{gathered} \frac{\left(x^2+12x-20\right)-\left(12x-20\right)}{12x-20}=\frac{\left(x^2+8x+12\right)-\left(8x+12\right)}{8x+12} \\ \Rightarrow \frac{x^2+12x-20-12x+20}{12x-20}=\frac{x^2+8x+12-8x-12}{8x+12} \\ \Rightarrow \frac{x^2}{12x-20}=\frac{x^2}{8x+12} \end{gathered}$$
This equation is true for x = 0. Therefore, x = 0 is a solution of the given equation.

If x ≠ 0, then x² ≠ 0.

Dividing both sides by x², we get

$$\begin{gathered} \frac{1}{12x-20}=\frac{1}{8x+12} \\ \Rightarrow 12x-20=8x+12 \\ \Rightarrow 12x-8x=20+12 \\ \Rightarrow 4x=32 \\ \Rightarrow x=8 \end{gathered}$$
Thus, the solutions of the given equation are x = 0 and x = 8.

(ii)
$$\begin{gathered} \frac{10x^2+15x+63}{5x^2-25x+12}=\frac{2x+3}{x-5} \\ \Rightarrow \frac{10x^2+15x+63}{2x+3}=\frac{5x^2-25x+12}{x-5} \end{gathered}$$
Multiplying both sides by $\frac{1}{5x}$, we get

$\frac{10x^2+15x+63}{10x^2+15x}=\frac{5x^2-25x+12}{5x^2-25x}$

Using dividendo, we get

$$\begin{gathered} \frac{\left(10x^2+15x+63\right)-\left(10x^2+15x\right)}{10x^2+15x}=\frac{\left(5x^2-25x+12\right)-\left(5x^2-25x\right)}{5x^2-25x} \\ \Rightarrow \frac{63}{10x^2+15x}=\frac{12}{5x^2-25x} \\ \Rightarrow \frac{63}{5x\left(2x+3\right)}=\frac{12}{5x\left(x-5\right)} \\ \Rightarrow \frac{63}{2x+3}=\frac{12}{x-5} \end{gathered}$$
$$\begin{gathered} \Rightarrow 63x-315=24x+36 \\ \Rightarrow 63x-24x=315+36 \\ \Rightarrow 39x=351 \\ \Rightarrow x=\frac{351}{39}=9 \end{gathered}$$
Thus, the solution of the give equation is x = 9.

(iii)
$\frac{{\left(2x+1\right)}^2+{\left(2x-1\right)}^2}{{\left(2x+1\right)}^2-{\left(2x-1\right)}^2}=\frac{17}{8}$

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{\left[{\left(2x+1\right)}^2+{\left(2x-1\right)}^2\right]+\left[{\left(2x+1\right)}^2-{\left(2x-1\right)}^2\right]}{\left[{\left(2x+1\right)}^2+{\left(2x-1\right)}^2\right]-\left[{\left(2x+1\right)}^2-{\left(2x-1\right)}^2\right]}=\frac{17+8}{17-8} \\ \Rightarrow \frac{2{\left(2x+1\right)}^2}{2{\left(2x-1\right)}^2}=\frac{25}{9} \\ \Rightarrow \frac{2x+1}{2x-1}=\sqrt{\frac{25}{9}}=\frac{5}{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow 6x+3=10x-5 \\ \Rightarrow 10x-6x=5+3 \\ \Rightarrow 4x=8 \\ \Rightarrow x=2 \end{gathered}$$
Thus, the solution of the given equation is x = 2.

(iv)
$\frac{\sqrt{4x+1}+\sqrt{x+3}}{\sqrt{4x+1}-\sqrt{x+3}}=\frac{4}{1}$

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{\left(\sqrt{4x+1}+\sqrt{x+3}\right)+\left(\sqrt{4x+1}-\sqrt{x+3}\right)}{\left(\sqrt{4x+1}+\sqrt{x+3}\right)-\left(\sqrt{4x+1}-\sqrt{x+3}\right)}=\frac{4+1}{4-1} \\ \Rightarrow \frac{2\sqrt{4x+1}}{2\sqrt{x+3}}=\frac{5}{3} \\ \Rightarrow \frac{\sqrt{4x+1}}{\sqrt{x+3}}=\frac{5}{3} \end{gathered}$$
Squaring on both sides, we get

$$\begin{gathered} \frac{4x+1}{x+3}={\left(\frac{5}{3}\right)}^2=\frac{25}{9} \\ \Rightarrow 36x+9=25x+75 \\ \Rightarrow 36x-25x=75-9 \\ \Rightarrow 11x=66 \\ \Rightarrow x=6 \end{gathered}$$
Thus, the solution of the given equation is x = 6.

(v)
$$\begin{gathered} \frac{{\left(4x+1\right)}^2+{\left(2x+3\right)}^2}{4x^2+12x+9}=\frac{61}{36} \\ \Rightarrow \frac{{\left(4x+1\right)}^2+{\left(2x+3\right)}^2}{{\left(2x+3\right)}^2}=\frac{61}{36} \end{gathered}$$
Applying dividendo, we get

$$\begin{gathered} \frac{{\left(4x+1\right)}^2+{\left(2x+3\right)}^2-{\left(2x+3\right)}^2}{{\left(2x+3\right)}^2}=\frac{61-36}{36} \\ \Rightarrow \frac{{\left(4x+1\right)}^2}{{\left(2x+3\right)}^2}=\frac{25}{36} \end{gathered}$$
Taking square root on both sides, we get

$$\begin{gathered} \frac{4x+1}{2x+3}=\sqrt{\frac{25}{36}}=\frac{5}{6} \\ \Rightarrow 24x+6=10x+15 \\ \Rightarrow 24x-10x=15-6 \\ \Rightarrow 14x=9 \\ \Rightarrow x=\frac{9}{14} \end{gathered}$$
Thus, the solution of the given equation is $x=\frac{9}{14}$.

(vi)
$\frac{{\left(3x-4\right)}^3-{\left(x+1\right)}^3}{{\left(3x-4\right)}^3+{\left(x+1\right)}^3}=\frac{61}{189}$

Applying componendo and dividendo, we get

$$\begin{gathered} \frac{\left[{\left(3x-4\right)}^3-{\left(x+1\right)}^3\right]+\left[{\left(3x-4\right)}^3+{\left(x+1\right)}^3\right]}{\left[{\left(3x-4\right)}^3-{\left(x+1\right)}^3\right]-\left[{\left(3x-4\right)}^3+{\left(x+1\right)}^3\right]}=\frac{61+189}{61-189} \\ \Rightarrow \frac{2{\left(3x-4\right)}^3}{-2{\left(x+1\right)}^3}=\frac{250}{-128} \\ \Rightarrow \frac{{\left(3x-4\right)}^3}{{\left(x+1\right)}^3}=\frac{125}{64} \end{gathered}$$
Taking cube root on both sides, we get

$$\begin{gathered} \frac{3x-4}{x+1}=\sqrt[3]{\frac{125}{64}} \\ \Rightarrow \frac{3x-4}{x+1}=\sqrt[3]{{\left(\frac{{\displaystyle 5}}{{\displaystyle 4}}\right)}^3}=\frac{5}{4} \\ \Rightarrow 12x-16=5x+5 \\ \Rightarrow 12x-5x=16+5 \\ \Rightarrow 7x=21 \\ \Rightarrow x=3 \end{gathered}$$
Thus, the solution of the given equation is x = 3.