Quadratic Equations: Question 3

Worked solution

Method: completing the square

Start with $x^2 - 5x - 3 = 0.$

Step 1: halve the coefficient of $x$. The coefficient of $x$ is $-5$, and half of it is $-\tfrac52$. So the square we need is $\left(x - \tfrac52\right)^2 = x^2 - 5x + \tfrac{25}{4}.$

Step 2: rewrite $x^2 - 5x$ using this square. Since $\left(x-\tfrac52\right)^2$ contains an extra $+\tfrac{25}{4}$, we must subtract it back: $x^2 - 5x = \left(x - \tfrac52\right)^2 - \tfrac{25}{4}.$

Step 3: substitute into the equation. $\left(x - \tfrac52\right)^2 - \tfrac{25}{4} - 3 = 0.$

Step 4: collect the constants (note $-3 = -\tfrac{12}{4}$): $\left(x - \tfrac52\right)^2 = \tfrac{25}{4} + 3 = \tfrac{25}{4} + \tfrac{12}{4} = \tfrac{37}{4}.$

Step 5: take the square root (remember the $\pm$): $x - \tfrac52 = \pm\sqrt{\tfrac{37}{4}} = \pm\frac{\sqrt{37}}{2}.$

Step 6: solve for $x$. $x = \frac52 \pm \frac{\sqrt{37}}{2} = \frac{5 \pm \sqrt{37}}{2}.$

Numerically, $\sqrt{37}\approx 6.08$, so $x \approx 5.54$ or $x \approx -0.54$.

Alternative method: the quadratic formula

With $a=1,\ b=-5,\ c=-3$: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-3)}}{2} = \frac{5 \pm \sqrt{25 + 12}}{2} = \frac{5 \pm \sqrt{37}}{2}.$

This confirms the completing-the-square result.

Checking the distractors

• B $\left(\dfrac{-5\pm\sqrt{37}}{2}\right)$: comes from using $\left(x+\tfrac52\right)^2$, a sign error in the bracket.
• C $\left(5\pm 2\sqrt{7}\right)$: comes from forgetting to halve $5$, using $(x-5)^2$ and getting $(x-5)^2=28$.
• D $\left(\dfrac{5\pm\sqrt{13}}{2}\right)$: comes from $\tfrac{25}{4}-3=\tfrac{13}{4}$, a sign slip on the constant.

Quick verification of A: for $x=\dfrac{5+\sqrt{37}}{2}$, the sum of roots is $\dfrac{5+\sqrt{37}}{2}+\dfrac{5-\sqrt{37}}{2}=5$ (matches $-b/a$) and the product is $\dfrac{25-37}{4}=-3$ (matches $c/a$). Correct.

$\boxed{x = \dfrac{5 \pm \sqrt{37}}{2}}$