Quadratics: Question 2

Worked solution

Part (a): Completing the square

Factor $3$ out of only the terms containing $x$:

$3x^2 - 12x + 7 = 3(x^2 - 4x) + 7$

Complete the square inside the bracket, using $x^2-4x = (x-2)^2 - 4$:

$= 3\big[(x-2)^2 - 4\big] + 7$

Multiply through by the $3$, then combine the constants:

$= 3(x-2)^2 - 12 + 7 = 3(x-2)^2 - 5$

So $f(x) = 3(x-2)^2 - 5$, giving $a = 3$, $p = 2$, $q = -5$.

Part (b): Minimum point

In the form $a(x-p)^2+q$, the vertex of the parabola is at $(p, q) = (2, -5)$.

Since $a = 3 > 0$, the graph is an upward-opening parabola (a "$\cup$" shape), so this vertex is the lowest point on the curve, a minimum, not a maximum.

Part (c): Solving $f(x) = 0$ exactly

Using the completed-square form:

$3(x-2)^2 - 5 = 0$

$(x-2)^2 = \frac{5}{3}$

$x - 2 = \pm\sqrt{\frac{5}{3}} = \pm\frac{\sqrt{5}}{\sqrt{3}} = \pm\frac{\sqrt{15}}{3}$

$x = 2 \pm \frac{\sqrt{15}}{3} = \frac{6 \pm \sqrt{15}}{3}$

Check using the quadratic formula. With $a=3$, $b=-12$, $c=7$:

$\text{discriminant} = (-12)^2 - 4(3)(7) = 144 - 84 = 60$

$x = \frac{12 \pm \sqrt{60}}{6} = \frac{12 \pm 2\sqrt{15}}{6} = \frac{6 \pm \sqrt{15}}{3}$

This agrees exactly with the completed-square method.

Final answers

• (a) $f(x) = 3(x-2)^2 - 5$, with $a=3$, $p=2$, $q=-5$
• (b) Minimum point $\boxed{(2, -5)}$
• (c) $x = \boxed{\dfrac{6 \pm \sqrt{15}}{3}}$