Question

Find the real solutions of each equation. $$ 2 x^{4}-5 x^{2}-12=0 $$

Short answer

The real solutions are \(x = 2\) and \(x = -2\).

Step-by-step solution

Recognize the Equation Type

Observe that the equation is a quadratic in terms of \(x^2\). We can rewrite it as \(2y^2 - 5y - 12 = 0\) by letting \(y = x^2\).

Solve the Quadratic Equation

We need to solve the quadratic equation \(2y^2 - 5y - 12 = 0\). We use the quadratic formula, \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] with \(a = 2\), \(b = -5\), and \(c = -12\).

Apply the Quadratic Formula

Substitute the values into the quadratic formula: \[ y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-12)}}{2(2)} \] Simplifying inside the square root: \[ y = \frac{5 \pm \sqrt{25 + 96}}{4} \] \[ y = \frac{5 \pm \sqrt{121}}{4} \] \[ y = \frac{5 \pm 11}{4} \]

Find the Values of \(y\)

Calculate the possible values for \(y\): \[ y = \frac{5 + 11}{4} = 4 \] \[ y = \frac{5 - 11}{4} = -1.5 \] So the solutions for \(y\) are \(y = 4\) and \(y = -1.5\).

Re-substitute \(x^2\) for \(y\)

Recall that \(y = x^2\). Therefore, \(x^2 = 4\) or \(x^2 = -1.5\). Only the solution \(x^2 = 4\) gives real numbers, since the square root of a negative number is not a real number.

Solve for \(x\)

Solve \(x^2 = 4\): \[ x = \pm \sqrt{4} \] \[ x = \pm 2 \] So, the real solutions are \(x = 2\) and \(x = -2\).

Key concepts

Solving Quadratic Equations

To solve a quadratic equation, we start by identifying its standard form: