Quadratic Equations

Half of $6$ is $3$, and $3^2 = 9$: $x^2 + 6x = (x + 3)^2 - 9$, so $a = 3$, $b = -9$. ($a = 6$ used the full coefficient; $b = 9$ added instead of subtracting.)

Let the integers be $n$ and $n + 1$: $n(n + 1) = 56$ gives $n^2 + n - 56 = 0$, so $(n + 8)(n - 7) = 0$. Since the integers are positive, $n = 7$ (the numbers are $7$ and $8$).

$(x + 3)^2 - 9 = -5$, so $(x + 3)^2 = 4$ and $x + 3 = \pm 2$, giving $x = -1$ or $x = -5$. (Don't drop the $\pm$.)

$x^2 - 8x + 20 = (x - 4)^2 - 16 + 20 = (x - 4)^2 + 4$, so the vertex is $(4, 4)$. ($(4, 20)$ read the constant term instead of completing the square.)

Let the length and width be $L$ and $W$. The perimeter gives $L + W = 13$ and the area gives $LW = 40$. So $L$ and $W$ are the roots of $x^2 - 13x + 40 = 0$, which factorises as $(x - 5)(x - 8) = 0$: the dimensions are $5$ cm and $8$ cm.

$x = \dfrac{-4 \pm \sqrt{16 + 8}}{4} = \dfrac{-4 \pm \sqrt{24}}{4} = \dfrac{-4 \pm 2\sqrt{6}}{4} = \dfrac{-2 \pm \sqrt{6}}{2}$ (divide top and bottom by 2). (Forgetting to simplify $\sqrt{24} = 2\sqrt{6}$ leaves an unsimplified answer.)

Set the $y$ values equal: $x^2 - 3x + 4 = x + 1$, so $x^2 - 4x + 3 = 0$ and $(x - 1)(x - 3) = 0$, giving $x = 1$ or $x = 3$.

$\Delta = 49 - 24 = 25$, so $x = \dfrac{7 \pm 5}{4}$, giving $x = 3$ or $x = \dfrac{1}{2}$. (The denominator is $2a = 4$.)

$\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16 < 0$, so there are no real solutions. ($\Delta > 0$ would give two; $\Delta = 0$ one.)

A tangent meets the curve exactly once, so $x^2 + 4 = 2x + c$, i.e. $x^2 - 2x + (4 - c) = 0$, must have $\Delta = 0$. Then $(-2)^2 - 4(1)(4 - c) = 0$, so $4 - 16 + 4c = 0$ and $c = 3$.

The maximum is at the vertex, $t = -\dfrac{b}{2a} = -\dfrac{30}{2(-5)} = 3$ s. Then $h = -5(3)^2 + 30(3) = -45 + 90 = 45$ m.

Factor the leading coefficient first: $2(x^2 + 4x) + 3 = 2((x + 2)^2 - 4) + 3 = 2(x + 2)^2 - 8 + 3 = 2(x + 2)^2 - 5$. ($2(x + 4)^2 - 29$ halved the full coefficient instead of halving after factoring.)

$\Delta > 0$ gives a positive square root, so the $\pm$ yields two distinct real roots (the parabola crosses the x-axis twice). ($\Delta = 0$ gives one repeated root.)

Two distinct real roots need $\Delta > 0$: $\Delta = k^2 - 4(1)(9) = k^2 - 36 > 0$, so $k^2 > 36$, giving $k > 6 \text{ or } k < -6$. ($-6 < k < 6$ solves $k^2 < 36$ (the no-real-roots case); $k = \pm 6$ gives equal roots.)

$\Delta = 36 - 4k = 0$, so $4k = 36$ and $k = 9$. ($36$ forgot to divide by $4$.) Then $x^2 + 6x + 9 = (x + 3)^2$.

The sum of the roots is $-\dfrac{b}{a} = -\dfrac{-12}{3} = 4$. ($\dfrac{7}{3}$ is the product $\dfrac{c}{a}$; watch the sign of $b$.)

A monic quadratic with roots summing to $S$ and multiplying to $P$ is $x^2 - Sx + P = 0$. Here $x^2 - 5x + 6 = 0$ (which factorises as $(x - 2)(x - 3) = 0$). (Note the sign: the sum enters with a minus.)

Product of roots $= \dfrac{c}{a} = \dfrac{-5}{2} = -\dfrac{5}{2}$. (The sum would be $-\dfrac{b}{a} = \dfrac{3}{2}$; $-5$ forgot to divide by $a = 2$.)

$y = x^2 - 9$ crosses the x-axis at $x = \pm 3$ and dips below between the roots, so $x^2 - 9 < 0$ gives $-3 < x < 3$. ($x^2 > 9$ would give the two outer regions.)

Multiply every term by $x$: $x^2 + 6 = 5x$, so $x^2 - 5x + 6 = 0$ and $(x - 2)(x - 3) = 0$, giving $x = 2$ or $x = 3$.

$\Delta = 16 - 4 = 12$, $\sqrt{12} = 2\sqrt{3}$, so $x = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$. ($x = 2 \pm 2\sqrt{3}$ forgot to simplify before dividing.)

$x(x + 3) = 28$ gives $x^2 + 3x - 28 = 0$, so $(x + 7)(x - 4) = 0$ and $x = -7$ or $x = 4$. A width must be positive, so $x = 4$.

$-5t^2 + 20t = 0$ gives $-5t(t - 4) = 0$, so $t = 0$ (thrown) or $t = 4$ (lands). The ball hits the ground at $t = 4$ s. ($t = 2$ is the peak.)

The outer rectangle is $(8 + 2x)$ by $(6 + 2x)$, so $(8 + 2x)(6 + 2x) = 120$. Expanding gives $4x^2 + 28x + 48 = 120$, so $4x^2 + 28x - 72 = 0$, i.e. $x^2 + 7x - 18 = 0$ and $(x + 9)(x - 2) = 0$. A width must be positive, so $x = 2$ cm.

Let the shorter leg be $x$, so the other is $x + 3$. By Pythagoras, $x^2 + (x + 3)^2 = 15^2$, giving $2x^2 + 6x + 9 = 225$, so $2x^2 + 6x - 216 = 0$, i.e. $x^2 + 3x - 108 = 0$ and $(x - 9)(x + 12) = 0$. A length is positive, so $x = 9$ cm (a $9, 12, 15$ triangle).