The powerful method of substitution is used to solve different types of equations.
Examples with Solutions
Example 1: Solve \(x - 3\sqrt{x} = -2\)
- Let \(u = \sqrt{x}\), so that \(u^2 = x\). Substituting gives: \[ u^2 - 3u = -2 \]
- Rewrite as a quadratic equation: \[ u^2 - 3u + 2 = 0 \]
- Solve for \(u\): \[ u = 1 \quad \text{or} \quad u = 2 \]
- Substitute back \(u = \sqrt{x}\) and solve for \(x\): \[ \sqrt{x} = 1 \quad \Rightarrow \quad x = 1 \] \[ \sqrt{x} = 2 \quad \Rightarrow \quad x = 4 \]
Example 2: Solve \(\dfrac{1}{(x-1)^2} - \dfrac{1}{x-1} - 2 = 0\)
- Let \(u = \dfrac{1}{x-1}\). Substituting gives: \[ u^2 - u - 2 = 0 \]
- Solve for \(u\): \[ u = -1 \quad \text{or} \quad u = 2 \]
- Substitute back and solve for \(x\): \[ \dfrac{1}{x-1} = -1 \quad \Rightarrow \quad x = 0 \] \[ \dfrac{1}{x-1} = 2 \quad \Rightarrow \quad x = \dfrac{3}{2} \]
Example 3: Solve \(-(x+3)^6 + 4(x+3)^3 = -21\)
- Let \(u = (x+3)^3\). Substituting gives: \[ -u^2 + 4u = -21 \quad \Rightarrow \quad u^2 - 4u - 21 = 0 \]
- Solve for \(u\): \[ u = -3 \quad \text{or} \quad u = 7 \]
- Substitute back and solve for \(x\): \[ (x+3)^3 = -3 \quad \Rightarrow \quad x = -3 - \sqrt[3]{3} \] \[ (x+3)^3 = 7 \quad \Rightarrow \quad x = -3 + \sqrt[3]{7} \]
Example 4: Solve \(3e^{2x} - e^x - 2 = 0\)
- Let \(u = e^x\), then \(u^2 = e^{2x}\). Substituting gives: \[ 3u^2 - u - 2 = 0 \]
- Solve for \(u\): \[ u = 1 \quad \text{or} \quad u = -\dfrac{2}{3} \]
- Substitute back \(u = e^x\) and solve for \(x\): \[ e^x = 1 \quad \Rightarrow \quad x = 0 \] \[ e^x = -\dfrac{2}{3} \quad \text{has no solution since } e^x > 0 \]
Example 5: Solve \(\sin^2 x - 4\sin x - 5 = 0\) on \([0, 2\pi)\)
- Let \(u = \sin x\). Substituting gives: \[ u^2 - 4u - 5 = 0 \]
- Solve for \(u\): \[ u = -1 \quad \text{or} \quad u = 5 \]
- Substitute back \(\sin x = u\) and solve: \[ \sin x = -1 \quad \Rightarrow \quad x = \dfrac{3\pi}{2} \] \[ \sin x = 5 \quad \text{has no solution since } -1 \le \sin x \le 1 \]