Quadratic equations

Answer: $8$

Half the coefficient of $x$ is $-4\mathclose{,}$ so $x^{2} - 8 x + 24 = \left(x - 4\right)^{2} - (-4)^{2} + 24 = \left(x - 4\right)^{2} + 8\mathclose{.}$ So $q = 8\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the square of half the coefficient instead of subtracting it → $40$
• Subtracted the square of the WHOLE coefficient → $-40$

Answer: $y = 3 x^{2} - 24 x + 47$

In that order: the stretch gives $y = 3x^{2}\mathclose{,}$ the shifts then give $y = 3\left(x - 4\right)^{2} - 1\mathclose{,}$ which is $y = 3 x^{2} - 24 x + 47\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 3 x^{2} - 24 x + 45$

Answer: $-3$

Half the coefficient of $x$ is $5\mathclose{;}$ add $5^{2} = 25$ to BOTH sides: $\left(x + 5\right)^{2} = 4\mathclose{.}$ So $x + 5 = \pm 2\mathclose{,}$ giving $x = -7$ or $x = -3\mathclose{;}$ the larger is $-3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to subtract $\frac{b}{2}$ at the end → $2$

Answer: $4$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} - 5 x + 4 = 0\mathclose{,}$ which factorises as $\left(x - 1\right)\left(x - 4\right) = 0\mathclose{.}$ The solutions are $1$ and $4\mathclose{;}$ the larger is $4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $1$
• Read the solutions off the brackets with the signs unchanged → $-1$

Answer: $-8$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}\mathclose{.}$ Here $a = -4$ and $b = 24\mathclose{,}$ so $x = -\dfrac{24}{2 \cdot (-4)} = 3\mathclose{.}$ The maximum value is $f(3) = -8\mathclose{.}$ (Completing the square gives $f(x) = -4\left(x - 3\right)^{2} - 8\mathclose{,}$ which shows the same thing.)

Common mistakes, and the answer each one gives:

• Gave the $x$-coordinate of the turning point instead of the value → $3$
• Substituted $x = \dfrac{b}{2a}\mathclose{,}$ without the minus sign → $-152$

Answer: $y = 2 x^{2} - 4 x - 1$

Shifting 1 unit to the right replaces $x$ by $x - 1\mathclose{;}$ shifting 3 units down adds $-3\mathclose{:}$ $y = 2\left(x - 1\right)^{2} - 3\mathclose{,}$ which is $y = 2 x^{2} - 4 x - 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the wrong way horizontally → $y = 2 x^{2} + 4 x - 1$
• Shifted the wrong way vertically → $y = 2 x^{2} - 4 x + 5$

Answer: $5$

Factorise: $x^{2} - 4 x - 5 = \left(x - 5\right)\left(x + 1\right) = 0\mathclose{,}$ so $x = 5$ or $x = -1\mathclose{.}$ The larger is $5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $-1$
• Read the solutions off the brackets with the signs unchanged → $1$

Answer: $-4 + \sqrt{21}$

Here $a = 1\mathclose{,}$ $b = 8$ and $c = -5\mathclose{,}$ so the discriminant is $b^2 - 4ac = 64 - (-20) = 84$ — positive but not a perfect square, so the formula is needed: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-8 \pm \sqrt{84}}{2}\mathclose{.}$ Since $\sqrt{84} = 2 \sqrt{21}\mathclose{,}$ that is $\dfrac{-8 \pm 2 \sqrt{21}}{2} = -4 \pm \sqrt{21}\mathclose{.}$ The larger solution is $-4 + \sqrt{21}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $4 + \sqrt{21}$
• Divided by $a$ instead of $2a$ → $-8 + 2\sqrt{21}$
• Gave the smaller solution → $-4 - \sqrt{21}$

Answer: $2$

The discriminant decides: $D = b^2 - 4ac = (-1)^{2} - 4 \cdot 1 \cdot (-4) = 17\mathclose{.}$ Positive: two solutions.

Common mistakes, and the answer each one gives:

• Used $b^2 + 4ac$ instead of $b^2 - 4ac$ → $0$

Answer: $x \le -5 \text{ or } x \ge 2$

Factorise: $\left(x + 5\right)\left(x - 2\right) \ge 0\mathclose{.}$ The parabola opens upwards and cuts the axis at $-5$ and $2\mathclose{,}$ so it is above the axis outside them: $x \le -5 \text{ or } x \ge 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the region on the wrong side of the roots → $-5 \le x \le 2$
• Used the roots with their signs swapped → $x \le -2 \text{ or } x \ge 5$

Answer: $-1$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}\mathclose{.}$ Here $a = -4$ and $b = -8\mathclose{,}$ so $x = -\dfrac{-8}{2 \cdot (-4)} = -1\mathclose{.}$ (Completing the square gives $f(x) = -4\left(x + 1\right)^{2} + 3\mathclose{,}$ which shows the same thing.)

Common mistakes, and the answer each one gives:

• Used $\dfrac{b}{2a}$ without the minus sign → $1$
• Gave the maximum VALUE instead of where it happens → $3$

Answer: $(2, 4)$

The vertex starts at $(2, 2)\mathclose{.}$ Reflecting in the $x$-axis changes the sign of its $y$-coordinate, giving $(2, -2)\mathclose{;}$ shifting up adds $6\mathclose{:}$ $(2, 4)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $y$-axis instead → $(-2, 8)$
• Shifted up before reflecting → $(2, -8)$

Answer: $6$

Set the two expressions equal and bring everything to one side: $2 x^{2} - 18 x + 36 = 0\mathclose{.}$ The $x^2$ terms do NOT cancel here. Divide by $2$ and factorise: $x^{2} - 9 x + 18 = \left(x - 3\right)\left(x - 6\right) = 0\mathclose{,}$ so $x = 3$ or $x = 6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $3$
• Treated the $x^2$ terms as if they cancelled → $2$

Answer: $16$

Exactly one solution means $D = b^2 - 4ac = 0\mathclose{:}$ $(-8)^{2} - 4c = 0\mathclose{,}$ so $c = \dfrac{64}{4} = 16\mathclose{.}$ (Then the left side is $\left(x - 4\right)^{2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Left the 4 out: set $b^2 - c = 0$ → $64$
• Halved $b$ but did not square it → $4$

Answer: $-2$

Half the coefficient of $x$ is $-3\mathclose{,}$ so $x^{2} - 6 x + 7 = \left(x - 3\right)^{2} - (-3)^{2} + 7 = \left(x - 3\right)^{2} - 2\mathclose{.}$ So $q = -2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the square of half the coefficient instead of subtracting it → $16$
• Subtracted the square of the WHOLE coefficient → $-29$

Answer: $y = 3 x^{2} - 24 x + 43$

In that order: the stretch gives $y = 3x^{2}\mathclose{,}$ the shifts then give $y = 3\left(x - 4\right)^{2} - 5\mathclose{,}$ which is $y = 3 x^{2} - 24 x + 43\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 3 x^{2} - 24 x + 33$

Answer: $6$

Half the coefficient of $x$ is $-4\mathclose{;}$ add $-4^{2} = 16$ to BOTH sides: $\left(x - 4\right)^{2} = 4\mathclose{.}$ So $x - 4 = \pm 2\mathclose{,}$ giving $x = 6$ or $x = 2\mathclose{;}$ the larger is $6\mathclose{.}$

Common mistakes, and the answer each one gives:

• Forgot to subtract $\frac{b}{2}$ at the end → $2$

Answer: $7$

Where the graphs meet, the two expressions are equal. Bring everything to one side: $x^{2} - 2 x - 35 = 0\mathclose{,}$ which factorises as $\left(x + 5\right)\left(x - 7\right) = 0\mathclose{.}$ The solutions are $-5$ and $7\mathclose{;}$ the larger is $7\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $-5$
• Read the solutions off the brackets with the signs unchanged → $5$

Answer: $-8$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}\mathclose{.}$ Here $a = -4$ and $b = -16\mathclose{,}$ so $x = -\dfrac{-16}{2 \cdot (-4)} = -2\mathclose{.}$ The maximum value is $f(-2) = -8\mathclose{.}$ (Completing the square gives $f(x) = -4\left(x + 2\right)^{2} - 8\mathclose{,}$ which shows the same thing.)

Common mistakes, and the answer each one gives:

• Gave the $x$-coordinate of the turning point instead of the value → $-2$
• Substituted $x = \dfrac{b}{2a}\mathclose{,}$ without the minus sign → $-72$

Answer: $y = x^{2} - 10 x + 27$

Shifting 5 units to the right replaces $x$ by $x - 5\mathclose{;}$ shifting 2 units up adds $2\mathclose{:}$ $y = \left(x - 5\right)^{2} + 2\mathclose{,}$ which is $y = x^{2} - 10 x + 27\mathclose{.}$

Common mistakes, and the answer each one gives:

• Shifted the wrong way horizontally → $y = x^{2} + 10 x + 27$
• Shifted the wrong way vertically → $y = x^{2} - 10 x + 23$

Answer: $-4$

Factorise: $x^{2} + 13 x + 36 = \left(x + 4\right)\left(x + 9\right) = 0\mathclose{,}$ so $x = -4$ or $x = -9\mathclose{.}$ The larger is $-4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $-9$
• Read the solutions off the brackets with the signs unchanged → $9$

Answer: $\dfrac{-2 + \sqrt{13}}{3}$

Here $a = 3\mathclose{,}$ $b = 4$ and $c = -3\mathclose{,}$ so the discriminant is $b^2 - 4ac = 16 - (-36) = 52$ — positive but not a perfect square, so the formula is needed: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \dfrac{-4 \pm \sqrt{52}}{6}\mathclose{.}$ Since $\sqrt{52} = 2 \sqrt{13}\mathclose{,}$ that is $\dfrac{-4 \pm 2 \sqrt{13}}{6} = \dfrac{-2 \pm \sqrt{13}}{3}\mathclose{.}$ The larger solution is $\dfrac{-2 + \sqrt{13}}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+b$ in the numerator instead of $-b$ → $\dfrac{2 + \sqrt{13}}{3}$
• Divided by $a$ instead of $2a$ → $\dfrac{-4 + 2\sqrt{13}}{3}$
• Gave the smaller solution → $\dfrac{-2 - \sqrt{13}}{3}$

Answer: $0$

The discriminant decides: $D = b^2 - 4ac = (-6)^{2} - 4 \cdot 2 \cdot 5 = -4\mathclose{.}$ Negative: no real solutions.

Common mistakes, and the answer each one gives:

• Used $b^2 + 4ac$ instead of $b^2 - 4ac$ → $2$

Answer: $-2 < x < 1$

Bring everything to one side and factorise: $\left(x + 2\right)\left(x - 1\right) < 0\mathclose{.}$ The parabola opens upwards and cuts the axis at $-2$ and $1\mathclose{,}$ so it is below the axis between them: $-2 < x < 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the region on the wrong side of the roots → $x < -2 \text{ or } x > 1$
• Used the roots with their signs swapped → $-1 < x < 2$

Answer: $6$

The turning point of $ax^2+bx+c$ sits at $x = -\dfrac{b}{2a}\mathclose{.}$ Here $a = -4$ and $b = 48\mathclose{,}$ so $x = -\dfrac{48}{2 \cdot (-4)} = 6\mathclose{.}$ (Completing the square gives $f(x) = -4\left(x - 6\right)^{2} - 3\mathclose{,}$ which shows the same thing.)

Common mistakes, and the answer each one gives:

• Used $\dfrac{b}{2a}$ without the minus sign → $-6$
• Gave the maximum VALUE instead of where it happens → $-3$

Answer: $(5, 3)$

The vertex starts at $(5, 1)\mathclose{.}$ Reflecting in the $x$-axis changes the sign of its $y$-coordinate, giving $(5, -1)\mathclose{;}$ shifting up adds $4\mathclose{:}$ $(5, 3)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Reflected in the $y$-axis instead → $(-5, 5)$
• Shifted up before reflecting → $(5, -5)$

Answer: $4$

Set the two expressions equal and bring everything to one side: $2 x^{2} - 12 x + 16 = 0\mathclose{.}$ The $x^2$ terms do NOT cancel here. Divide by $2$ and factorise: $x^{2} - 6 x + 8 = \left(x - 2\right)\left(x - 4\right) = 0\mathclose{,}$ so $x = 2$ or $x = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the smaller solution → $2$
• Treated the $x^2$ terms as if they cancelled → $\dfrac{4}{3}$

Answer: $16$

Exactly one solution means $D = b^2 - 4ac = 0\mathclose{:}$ $8^{2} - 4c = 0\mathclose{,}$ so $c = \dfrac{64}{4} = 16\mathclose{.}$ (Then the left side is $\left(x + 4\right)^{2}\mathclose{.})$

Common mistakes, and the answer each one gives:

• Left the 4 out: set $b^2 - c = 0$ → $64$
• Halved $b$ but did not square it → $4$

Answer: $-8$

Half the coefficient of $x$ is $1\mathclose{,}$ so $x^{2} + 2 x - 7 = \left(x + 1\right)^{2} - 1^{2} - 7 = \left(x + 1\right)^{2} - 8\mathclose{.}$ So $q = -8\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the square of half the coefficient instead of subtracting it → $-6$
• Subtracted the square of the WHOLE coefficient → $-11$

Answer: $y = 2 x^{2} + 16 x + 29$

In that order: the stretch gives $y = 2x^{2}\mathclose{,}$ the shifts then give $y = 2\left(x + 4\right)^{2} - 3\mathclose{,}$ which is $y = 2 x^{2} + 16 x + 29\mathclose{.}$

Common mistakes, and the answer each one gives:

• Stretched AFTER shifting up or down, so the vertical shift was stretched too → $y = 2 x^{2} + 16 x + 26$