Quadratic equations - Answer key

Fluency

$(x - 2)(x - 5) = 0$ : $x = 2$ or $x = 5$ .
$(x + 5)(x - 3) = 0$ : $x = -5$ or $x = 3$ .
$3x(x - 4) = 0$ : $x = 0$ or $x = 4$ .
$(x - 5)(x + 5) = 0$ : $x = 5$ or $x = -5$ .
$x = \dfrac{-4 \pm \sqrt{16 - 4}}{2} = \dfrac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$ .
$x = \dfrac{3 \pm \sqrt{9 + 8}}{4} = \dfrac{3 \pm \sqrt{17}}{4}$ .
$\Delta = 36 - 36 = 0$ . One repeated solution: $x = -3$ .
$\Delta = 1 - 24 = -23 < 0$ . No real solutions.
$x^2 + 8x = -12$ . $(x + 4)^2 = 16 - 12 = 4$ . $x + 4 = \pm 2$ . $x = -2$ or $x = -6$ .
$x^2 + 4x - 5 = 7$ , so $x^2 + 4x - 12 = 0$ . $(x + 6)(x - 2) = 0$ : $x = -6$ or $x = 2$ .

Reasoning

$(x + 4)(x - 2) = 32$ , $x^2 + 2x - 8 = 32$ , $x^2 + 2x - 40 = 0$ . $x = \dfrac{-2 \pm \sqrt{4 + 160}}{2} = \dfrac{-2 \pm \sqrt{164}}{2} = -1 \pm \sqrt{41}$ . Since $x > 2$ : $x = -1 + \sqrt{41} \approx 5.40$ cm.
$\Delta = 144 - 144 = 0$ . $(2x - 3)^2 = 0$ , so $x = \dfrac{3}{2}$ . There is one repeated solution because the parabola touches the $x$ -axis at exactly one point.
(a) $x^2 - 6x = -3$ . $(x - 3)^2 = 9 - 3 = 6$ . $x = 3 \pm \sqrt{6}$ . (b) $x = \dfrac{6 \pm \sqrt{36 - 12}}{2} = \dfrac{6 \pm \sqrt{24}}{2} = 3 \pm \sqrt{6}$ . Both methods give the same answer.
Two distinct real solutions requires $\Delta > 0$ : $16 - 4k > 0$ , so $k < 4$ .
$25t - 5t^2 = 20$ . $5t^2 - 25t + 20 = 0$ . $t^2 - 5t + 4 = 0$ . $(t - 1)(t - 4) = 0$ : $t = 1$ s or $t = 4$ s.
$\dfrac{6}{x} + x = 5$ . Multiply by $x$ : $6 + x^2 = 5x$ . $x^2 - 5x + 6 = 0$ . $(x - 2)(x - 3) = 0$ : $x = 2$ or $x = 3$ .
Let the number be $x$ . $x + \dfrac{1}{x} = \dfrac{10}{3}$ . Multiply by $3x$ : $3x^2 + 3 = 10x$ . $3x^2 - 10x + 3 = 0$ . $(3x - 1)(x - 3) = 0$ : $x = \dfrac{1}{3}$ or $x = 3$ .
(a) $\Delta = 4 - 20 = -16 < 0$ , so no real solutions. (b) $x^2 + 2x + 5 = (x + 1)^2 + 4$ . Since $(x+1)^2 \geq 0$ , the expression $\geq 4 > 0$ , so it can never equal zero.

Reasoning

$ax^2 + bx + c = 0$ . Divide by $a$ : $x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$ . Complete the square: $\left(x + \dfrac{b}{2a}\right)^2 = \dfrac{b^2}{4a^2} - \dfrac{c}{a} = \dfrac{b^2 - 4ac}{4a^2}$ . Square root: $x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}$ . Therefore $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .
Width $= x$ , length $= 80 - 2x$ . Area $= x(80 - 2x) = 80x - 2x^2$ . Axis of symmetry: $x = \dfrac{80}{4} = 20$ . Maximum area: $80(20) - 2(400) = 1600 - 800 = 800$ m $^2$ . Dimensions: $20$ m by $40$ m.
$(1, 3)$ : $2 + b + c = 3$ , so $b + c = 1$ . $(3, 7)$ : $18 + 3b + c = 7$ , so $3b + c = -11$ . Subtract: $2b = -12$ , $b = -6$ . $c = 1 - (-6) = 7$ .
$x^2 = mx + 1$ , so $x^2 - mx - 1 = 0$ . Exactly one intersection: $\Delta = 0$ . $m^2 + 4 = 0$ . Since $m^2 \geq 0$ , $m^2 + 4 \geq 4 > 0$ for all real $m$ . So $\Delta > 0$ always, meaning the line always intersects the parabola at two points — there is no value of $m$ giving exactly one intersection.

Reasoning

Let $u = \dfrac{x}{x+1}$ , so $\dfrac{x+1}{x} = \dfrac{1}{u}$ . The equation becomes $u + \dfrac{1}{u} = \dfrac{5}{2}$ . Multiply by $2u$ : $2u^2 + 2 = 5u$ , $2u^2 - 5u + 2 = 0$ , $(2u - 1)(u - 2) = 0$ , $u = \dfrac{1}{2}$ or $u = 2$ . If $u = \dfrac{1}{2}$ : $\dfrac{x}{x+1} = \dfrac{1}{2}$ , $2x = x + 1$ , $x = 1$ . If $u = 2$ : $\dfrac{x}{x+1} = 2$ , $x = 2x + 2$ , $x = -2$ . Solutions: $x = 1$ or $x = -2$ .
By Vieta’s formulas: $\alpha + \beta = 5$ and $\alpha\beta = 3$ . $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 25 - 6 = 19$ . $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{5}{3}$ .
$\Delta = (k+2)^2 - 4k = k^2 + 4k + 4 - 4k = k^2 + 4$ . For a repeated root: $\Delta = 0$ , so $k^2 + 4 = 0$ . Since $k^2 \geq 0$ , $k^2 + 4 \geq 4 > 0$ for all real $k$ . There is no real value of $k$ that gives a repeated root (restriction: $k \neq 0$ since the equation must be quadratic).
$15 + 20t - 5t^2 = 30$ : $5t^2 - 20t + 15 = 0$ , $t^2 - 4t + 3 = 0$ , $(t-1)(t-3) = 0$ : $t = 1$ or $t = 3$ s. Hits ground: $15 + 20t - 5t^2 = 0$ , $t^2 - 4t - 3 = 0$ , $t = \dfrac{4 \pm \sqrt{16 + 12}}{2} = 2 \pm \sqrt{7}$ . Taking the positive root: $t = 2 + \sqrt{7} \approx 4.65$ s.

Year 10 core - answers