Cracking Quadratic equation: Two Equations, One Real Challenge!

Find the solution set of each of the following equations in \(\mathbb{R}\):

1. \(X^2 - 2X - 6 = 0\)

2. \(X + \frac{5}{X} = 4\), \(X \neq 0\)

Solution

1. The expression: \(X^2 - 2X - 6\) is difficult to be factorized, so we use the general formula.

∴ a = 1, b = -2, c = -6

∴ \(X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(2) \pm \sqrt{(2)^2 - 4 \times 1 \times (-6)}}{2 \times 1}\)

\[= \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}\]

∴ The solution set = \(\left\{ 1 + \sqrt{7}, 1 - \sqrt{7} \right\}\)

2. ∵ \(X + \frac{5}{X} = 4\)

"By multiplying both sides of the equation by \(X\)"

∴ \(X^2 + 5 = 4X\)

∴ \(X^2 - 4X + 5 = 0\)

"Notice putting the equation in the form: \(aX^2 + bX + c = 0\)"

∴ a = 1, b = -4, c = 5

∴ \(X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{16 - 4 \times 1 \times 5}}{2} = \frac{4 \pm \sqrt{-4}}{2}\)

∴ \(\sqrt{-4} \notin \mathbb{R}\) ∴ There is no real roots of the equation: \(X^2 - 4X + 5 = 0\)

∴ The solution set = \(\emptyset\)

Created with Mr. Ayman Hassan for math Egypt lovers