1.4 Equations reducible to quadratic form

Example 1: Solve the equation $x^{4} - 13x^{2} + 36 = 0$

Solution: $x^{4} - 13x^{2} + 36 = 0$ (i)

Let $x^{2} = y$ . Then $x^{4} = y^{2}$

Equation (i) becomes $y^{2} - 13y + 36 = 0$ which can be factorized as $y^{2} - 9y - 4y + 36 = 0$ $y(y - 9) - 4(y - 9) = 0$ $(y - 9)(y - 4) = 0$

Either $y - 9 = 0$ or $y - 4 = 0$ , that is, $y = 9 \qquad \text{or} \qquad y = 4$

$\text{Put} y = x^{2}$ $x^{2} = 9 \qquad \text{or} \qquad x^{2} = 4$ $\Rightarrow x=\pm 3\text{or}x=\pm 2$$\Rightarrow x = \pm 3 \qquad \text{or} \qquad x = \pm 2$

The solution set is $\{\pm 2,\pm 3\}$$\{\pm 2, \pm 3\}$

Example 2: Solve the equation $2(2x-1)+\frac{3}{2x-1}=5$$2(2x - 1) + \frac{3}{2x - 1} = 5$

Solution: Given that $2(2x-1)+\frac{3}{2x-1}=5$$2(2x - 1) + \frac{3}{2x - 1} = 5$ (i)

Let $2x - 1 = y$

Then the equation (i) becomes $2y+\frac{3}{y}=5$$2y + \frac{3}{y} = 5$ or $2y^{2} + 3 = 5y$ $\Rightarrow 2y^2-5y+3=0$$\Rightarrow 2y^{2} - 5y + 3 = 0$

Using quadratic formula $y=\frac{-(-5)\pm \sqrt{(-5{)}^2-4\times 2\times 3}}{2\times 2}$$y = \frac{-(-5)\pm\sqrt{(-5)^{2} - 4\times2\times3}}{2\times2}$

​​y=2×2−(−5)±(−5)2−4×2×3

​​ $=\frac{5\pm \sqrt{25-24}}{4}=\frac{5\pm 1}{4}$$= \frac{5\pm\sqrt{25 - 24}}{4} = \frac{5\pm1}{4}$

​​=45±1​=45±25−24

​​=45±1​

We have $y=\frac{5+1}{4}=\frac{6}{4}=\frac{3}{2}$$y = \frac{5+1}{4} = \frac{6}{4} = \frac{3}{2}$ and $y=\frac{5-1}{4}=\frac{4}{4}=1$$y = \frac{5-1}{4} = \frac{4}{4} = 1$

When $y=\frac{3}{2}$$y = \frac{3}{2}$ , $2x-1=\frac{3}{2}$$2x - 1 = \frac{3}{2}$ $2x=\frac{3}{2}+1=\frac{5}{2}$$2x = \frac{3}{2} + 1 = \frac{5}{2}$ $\Rightarrow x=\frac{5}{4}$$\Rightarrow x = \frac{5}{4}$

When $y = 1$ , $2x - 1 = 1$ $2x = 1 + 1 = 2$ $\Rightarrow x=1$$\Rightarrow x = 1$

Thus, the solution set is $\{1,\frac{5}{4}\}$$\left\{1, \frac{5}{4}\right\}$