Quadratic Equations

Bi-quadratic equation reducible to quadratic form

Math

Examples

For example, solve the biquadratic equation: $12x^4 - 11x^2 + 2 = 0$

Explanation:

Step 1: Recognize that this is a biquadratic equation of the form $ax^4 + bx^2 + c = 0$ , where $a = 12, b = -11,$ and $c = 2$ .

Step 2: Substitute $y = x^2$ to obtain a quadratic equation in $y$ . This means we are using the substitution method to convert the biquadratic equation to a quadratic equation. Substituting $y = x^2$ gives:

$12y^2 - 11y + 2 = 0$

Step 3: Solve the quadratic equation using the quadratic formula:

$y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Plugging in $a = 12, b = -11,$ and $c = 2$ , we get:

$y = \frac{-(-11) ± \sqrt{(-11)^2 - 4(12)(2))}} { 2(12)}\\ y = \frac{-(-11) ± \sqrt{121- 96}} { 24}\\ y = \frac{-(-11) ± \sqrt{25}} { 24}\\</p><p>y= \frac{11 ± 5 }{24}\\y= \frac{16 }{24}; \space \space \frac{6 }{24} \\ or\\y= \frac{2 }{3}; \space \space \frac{1}{4}$

Therefore, $y= \frac{2 }{3}; \space \space \frac{1}{4}$

Step 4: Solve for $x$ using the values of $y$ found in Step 3. Remember that we substituted $y = x^2$ in Step 2, so we need to solve for $x$ using the values of $y$ found in Step 3. This gives:

$y = x^2\\ \\ x^2 = \frac23$ or $x^2 = \frac14$

Taking the square root of both sides, we get:

$x = ±\sqrt{\frac23}\space \space or \space \space x = ±\sqrt{\frac14}$

Simplifying the roots, we get:

$x = ±\frac{\sqrt{6}}3$ or $x = ±\frac12$

Therefore, the four solutions to the equation $12x^4 - 11x^2 + 2 = 0$ are $x = \frac{\sqrt{6}}3$ , $x = -\frac{\sqrt{6}}3$ , $x =\frac12$ , and $x = -\frac12$ .

Solution:

Given equation is $12x^4 -11x^2 + 2= 0$ , where $a = 12, b = -11,$ and $c = 2$ .

Substitute $y = x^2$ to obtain a quadratic equation in $y$ :

$12y^2 - 11y + 2 = 0$ Applying quadratic formula to above equation **** $y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$