The Quadratic Formula GCSE Questions

Answer $a = 3,\ b = -5,\ c = 1$

1. $\displaystyle 3x^2 - 5x + 1 = 0$Compare with $ax^2 + bx + c = 0$
2. $\displaystyle a = 3,\quad b = -5,\quad c = 1$Keep the sign with each number

• B1 All three correct, including $b = -5$

Answer $x = -3 \pm \sqrt{5}$

1. $\displaystyle a = 1,\quad b = 6,\quad c = 4$
2. $\displaystyle x = \frac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 4}}{2 \times 1}$Substitute into the formula
3. $\displaystyle = \frac{-6 \pm \sqrt{20}}{2}$$36 - 16 = 20$
4. $\displaystyle = \frac{-6 \pm 2\sqrt{5}}{2} = -3 \pm \sqrt{5}$$\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$

• M1 Correct substitution: $\dfrac{-6 \pm \sqrt{6^2 - 4 \times 1 \times 4}}{2 \times 1}$
• M1 $\dfrac{-6 \pm \sqrt{20}}{2}$
• A1 $-3 \pm \sqrt{5}$ — allow $\dfrac{-6 \pm \sqrt{20}}{2}$ for M1 M1 only

Answer $x = 4.65$ or $x = -0.65$

1. $\displaystyle a = 1,\quad b = -4,\quad c = -3$
2. $\displaystyle x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 1 \times (-3)}}{2 \times 1}$Use brackets around negative numbers
3. $\displaystyle = \frac{4 \pm \sqrt{28}}{2}$$16 + 12 = 28$: minus a negative is a plus
4. $\displaystyle x = 4.645\ldots \quad\text{or}\quad x = -0.645\ldots$
5. $\displaystyle x = 4.65 \quad\text{or}\quad x = -0.65$Round at the end

• M1 Correct substitution: $\dfrac{4 \pm \sqrt{(-4)^2 - 4 \times 1 \times (-3)}}{2}$
• M1 $\dfrac{4 \pm \sqrt{28}}{2}$
• A1 $4.65$ and $-0.65$

Answer $x = 0.28$ or $x = -1.78$

1. $\displaystyle a = 2,\quad b = 3,\quad c = -1$
2. $\displaystyle x = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-1)}}{2 \times 2}$
3. $\displaystyle = \frac{-3 \pm \sqrt{17}}{4}$$9 + 8 = 17$, and the bottom is $2a = 4$
4. $\displaystyle x = 0.280\ldots \quad\text{or}\quad x = -1.780\ldots$
5. $\displaystyle x = 0.28 \quad\text{or}\quad x = -1.78$

• M1 Correct substitution: $\dfrac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-1)}}{2 \times 2}$
• M1 $\dfrac{-3 \pm \sqrt{17}}{4}$
• A1 $0.28$ and $-1.78$

Answer $x = 1.45$ or $x = -3.45$

1. $\displaystyle a = 1,\quad b = 2,\quad c = -5$
2. $\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-5)}}{2 \times 1}$
3. $\displaystyle = \frac{-2 \pm \sqrt{24}}{2}$$4 + 20 = 24$
4. $\displaystyle x = 1.449\ldots \quad\text{or}\quad x = -3.449\ldots$
5. $\displaystyle x = 1.45 \quad\text{or}\quad x = -3.45$

• M1 Correct substitution into the formula
• M1 $\dfrac{-2 \pm \sqrt{24}}{2}$
• A1 $1.45$ and $-3.45$

Answer $x = 2.18$ or $x = 0.15$

1. $\displaystyle 3x^2 - 7x + 1 = 0$Rearrange so one side is $0$ first
2. $\displaystyle a = 3,\quad b = -7,\quad c = 1$
3. $\displaystyle x = \frac{7 \pm \sqrt{(-7)^2 - 4 \times 3 \times 1}}{2 \times 3} = \frac{7 \pm \sqrt{37}}{6}$
4. $\displaystyle x = 2.180\ldots \quad\text{or}\quad x = 0.152\ldots$
5. $\displaystyle x = 2.18 \quad\text{or}\quad x = 0.15$

• M1 $3x^2 - 7x + 1 = 0$
• M1 Correct substitution: $\dfrac{7 \pm \sqrt{49 - 12}}{6}$
• A1 $2.18$ and $0.15$

Answer $x = 3 \pm \sqrt{7}$

1. $\displaystyle x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 2}}{2}$$a = 1$, $b = -6$, $c = 2$
2. $\displaystyle = \frac{6 \pm \sqrt{28}}{2}$$36 - 8 = 28$
3. $\displaystyle = \frac{6 \pm 2\sqrt{7}}{2}$$\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$
4. $\displaystyle = 3 \pm \sqrt{7}$Divide both terms on top by $2$

• M1 Correct substitution into the formula
• M1 $\dfrac{6 \pm \sqrt{28}}{2}$
• A1 $3 \pm \sqrt{7}$

Answer $x = 0.72$ or $x = -1.12$

1. $\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4 \times 5 \times (-4)}}{2 \times 5}$
2. $\displaystyle = \frac{-2 \pm \sqrt{84}}{10}$$4 + 80 = 84$
3. $\displaystyle x = 0.716\ldots \quad\text{or}\quad x = -1.116\ldots$
4. $\displaystyle x = 0.72 \quad\text{or}\quad x = -1.12$

• M1 Correct substitution into the formula
• M1 $\dfrac{-2 \pm \sqrt{84}}{10}$
• A1 $0.72$ and $-1.12$

Answer $x = 1.32$ or $x = -5.32$

1. $\displaystyle x^2 + 4x - 7 = 0$Expand and make one side $0$
2. $\displaystyle x = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-7)}}{2} = \frac{-4 \pm \sqrt{44}}{2}$
3. $\displaystyle x = 1.316\ldots \quad\text{or}\quad x = -5.316\ldots$
4. $\displaystyle x = 1.32 \quad\text{or}\quad x = -5.32$

• M1 $x^2 + 4x - 7 = 0$
• M1 Correct substitution: $\dfrac{-4 \pm \sqrt{16 + 28}}{2}$
• A1 $1.32$ and $-5.32$

Answer $x = \dfrac{2 \pm \sqrt{10}}{2}$

1. $\displaystyle x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 2 \times (-3)}}{2 \times 2}$
2. $\displaystyle = \frac{4 \pm \sqrt{40}}{4}$$16 + 24 = 40$
3. $\displaystyle = \frac{4 \pm 2\sqrt{10}}{4}$$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$
4. $\displaystyle = \frac{2 \pm \sqrt{10}}{2}$Divide every term by $2$

• M1 Correct substitution: $\dfrac{4 \pm \sqrt{16 + 24}}{4}$
• M1 $\dfrac{4 \pm \sqrt{40}}{4}$
• A1 $\dfrac{2 \pm \sqrt{10}}{2}$

Answer $(x + 3)(x - 1) = 20$, so $x^2 + 2x - 3 = 20$, so $x^2 + 2x - 23 = 0$.

1. $\displaystyle (x + 3)(x - 1) = 20$Area $=$ length $\times$ width
2. $\displaystyle x^2 + 2x - 3 = 20$
3. $\displaystyle x^2 + 2x - 23 = 0$

• M1 $(x + 3)(x - 1) = 20$
• A1 Correct expansion and rearrangement to $x^2 + 2x - 23 = 0$

Answer $x = 3.90$

1. $\displaystyle x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-23)}}{2} = \frac{-2 \pm \sqrt{96}}{2}$
2. $\displaystyle x = 3.898\ldots \quad\text{or}\quad x = -5.898\ldots$
3. $\displaystyle x = 3.90$A negative $x$ would make the width negative, so reject it

• M1 Correct substitution: $\dfrac{-2 \pm \sqrt{4 + 92}}{2}$
• A1 $3.90$
• B1 Rejects the negative solution, because $x - 1$ would be negative

Answer $-b$ should be $+3$, and $-4ac = -4 \times 1 \times (-5) = +20$, so the square root is $\sqrt{29}$. $x = 4.19$ or $x = -1.19$.

1. $\displaystyle a = 1,\quad b = -3,\quad c = -5$
2. $\displaystyle -b = -(-3) = 3$Mistake 1: Hari used $-3$
3. $\displaystyle b^2 - 4ac = 9 - 4 \times 1 \times (-5) = 29$Mistake 2: minus a negative is a plus, so $+20$
4. $\displaystyle x = \frac{3 \pm \sqrt{29}}{2} = 4.192\ldots \text{ or } -1.192\ldots$
5. $\displaystyle x = 4.19 \quad\text{or}\quad x = -1.19$

• B1 Explains that $-b = +3$, because $b = -3$
• B1 Explains that $b^2 - 4ac = 9 + 20 = 29$, because $c$ is negative
• B1 $4.19$ and $-1.19$

Answer $x = \dfrac{3 \pm \sqrt{2}}{2}$

1. $\displaystyle x = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 4 \times 7}}{2 \times 4}$
2. $\displaystyle = \frac{12 \pm \sqrt{32}}{8}$$144 - 112 = 32$
3. $\displaystyle = \frac{12 \pm 4\sqrt{2}}{8}$$\sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$
4. $\displaystyle = \frac{3 \pm \sqrt{2}}{2}$Divide every term by $4$

• M1 Correct substitution: $\dfrac{12 \pm \sqrt{144 - 112}}{8}$
• M1 $\dfrac{12 \pm \sqrt{32}}{8}$ or $\dfrac{12 \pm 4\sqrt{2}}{8}$
• A1 $\dfrac{3 \pm \sqrt{2}}{2}$

Answer $2(x + 1) + x = x(x + 1)$, so $3x + 2 = x^2 + x$, so $x^2 - 2x - 2 = 0$.

1. $\displaystyle 2(x + 1) + x = x(x + 1)$Multiply every term by $x(x + 1)$
2. $\displaystyle 3x + 2 = x^2 + x$
3. $\displaystyle 0 = x^2 - 2x - 2$

• M1 Multiplies by $x(x + 1)$: $2(x + 1) + x = x(x + 1)$
• M1 $3x + 2 = x^2 + x$
• A1 Correct rearrangement to $x^2 - 2x - 2 = 0$

Answer $x = 1 \pm \sqrt{3}$

1. $\displaystyle x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-2)}}{2} = \frac{2 \pm \sqrt{12}}{2}$
2. $\displaystyle = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$$\sqrt{12} = 2\sqrt{3}$

• M1 $\dfrac{2 \pm \sqrt{4 + 8}}{2}$
• A1 $1 \pm \sqrt{3}$

Answer $t = 0.45$ s and $t = 3.55$ s

1. $\displaystyle 2 + 20t - 5t^2 = 10$Set the height equal to $10$
2. $\displaystyle 5t^2 - 20t + 8 = 0$
3. $\displaystyle t = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 5 \times 8}}{2 \times 5} = \frac{20 \pm \sqrt{240}}{10}$
4. $\displaystyle t = 0.450\ldots \quad\text{or}\quad t = 3.549\ldots$
5. $\displaystyle t = 0.45 \text{ s} \quad\text{and}\quad t = 3.55 \text{ s}$Once on the way up, once on the way down

• M1 $5t^2 - 20t + 8 = 0$ oe
• M1 Correct substitution into the formula
• A1 $0.45$
• A1 $3.55$

Answer $x = \dfrac{-8 \pm \sqrt{20}}{2} = \dfrac{-8 \pm 2\sqrt{5}}{2} = -4 \pm \sqrt{5}$

1. $\displaystyle x = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times 11}}{2}$
2. $\displaystyle = \frac{-8 \pm \sqrt{20}}{2}$$64 - 44 = 20$
3. $\displaystyle = \frac{-8 \pm 2\sqrt{5}}{2}$"Show that": every step must be written down
4. $\displaystyle = -4 \pm \sqrt{5}$

• M1 Correct substitution: $\dfrac{-8 \pm \sqrt{64 - 44}}{2}$
• M1 $\sqrt{20} = 2\sqrt{5}$ seen
• A1 Fully correct working to $-4 \pm \sqrt{5}$

Answer $x^2 + (x + 2)^2 = (2x - 1)^2$, so $2x^2 + 4x + 4 = 4x^2 - 4x + 1$, so $2x^2 - 8x - 3 = 0$.

1. $\displaystyle x^2 + (x + 2)^2 = (2x - 1)^2$Pythagoras: $a^2 + b^2 = c^2$
2. $\displaystyle x^2 + x^2 + 4x + 4 = 4x^2 - 4x + 1$
3. $\displaystyle 2x^2 + 4x + 4 = 4x^2 - 4x + 1$
4. $\displaystyle 0 = 2x^2 - 8x - 3$Move everything to the right-hand side

• M1 $x^2 + (x + 2)^2 = (2x - 1)^2$
• M1 Both sides expanded correctly: $2x^2 + 4x + 4 = 4x^2 - 4x + 1$
• A1 Correct rearrangement to $2x^2 - 8x - 3 = 0$

Answer $7.69$ cm

1. $\displaystyle x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 2 \times (-3)}}{2 \times 2} = \frac{8 \pm \sqrt{88}}{4}$
2. $\displaystyle x = 4.345\ldots \quad (\text{or } -0.345\ldots)$A length can't be negative
3. $\displaystyle 2x - 1 = 2 \times 4.345\ldots - 1 = 7.690\ldots$Use the unrounded value of $x$
4. $\displaystyle \text{hypotenuse} = 7.69 \text{ cm}$

• M1 Correct substitution: $\dfrac{8 \pm \sqrt{64 + 24}}{4}$
• A1 $x = 4.345\ldots$, rejecting the negative solution
• A1 $7.69$ cm

Answer $p = 6$

1. $\displaystyle x = \frac{-p \pm \sqrt{p^2 - 28}}{2}$The formula with $a = 1$, $b = p$, $c = 7$
2. $\displaystyle -\frac{p}{2} = -3$Match the part outside the square root with $-3$
3. $\displaystyle p = 6$
4. $\displaystyle \frac{\sqrt{36 - 28}}{2} = \frac{\sqrt{8}}{2}$Check: the surd part should match $\sqrt{2}$ too
5. $\displaystyle = \frac{2\sqrt{2}}{2} = \sqrt{2}$

• M1 $x = \dfrac{-p \pm \sqrt{p^2 - 28}}{2}$, or $\big(x - (-3 + \sqrt{2})\big)\big(x - (-3 - \sqrt{2})\big)$
• M1 $-\dfrac{p}{2} = -3$, or expands to $(x + 3)^2 - 2$
• A1 $p = 6$

Answer $x = 3.19$ or $x = -2.19$

1. $\displaystyle 2x + 1 = (x + 3)(x - 2)$Multiply both sides by $(x - 2)$
2. $\displaystyle 2x + 1 = x^2 + x - 6$
3. $\displaystyle x^2 - x - 7 = 0$
4. $\displaystyle x = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 1 \times (-7)}}{2} = \frac{1 \pm \sqrt{29}}{2}$
5. $\displaystyle x = 3.19 \quad\text{or}\quad x = -2.19$

• M1 $2x + 1 = (x + 3)(x - 2)$
• M1 $x^2 - x - 7 = 0$
• M1 Correct substitution: $\dfrac{1 \pm \sqrt{1 + 28}}{2}$ — ft their quadratic
• A1 $3.19$ and $-2.19$

Answer $5.41$ cm

1. $\displaystyle x^2 + (x + 3)^2 = 10^2$Let the width be $x$. The diagonal makes a right-angled triangle
2. $\displaystyle x^2 + x^2 + 6x + 9 = 100$
3. $\displaystyle 2x^2 + 6x - 91 = 0$
4. $\displaystyle x = \frac{-6 \pm \sqrt{6^2 - 4(2)(-91)}}{2(2)}$$4(2)(-91)$ means $4 \times 2 \times (-91)$
5. $\displaystyle = \frac{-6 \pm \sqrt{764}}{4}$$36 + 728 = 764$
6. $\displaystyle x = 5.410\ldots$Reject the negative solution: a width can't be negative
7. $\displaystyle \text{width} = 5.41 \text{ cm}$

• M1 $x^2 + (x + 3)^2 = 10^2$, where $x$ is the width
• M1 $2x^2 + 6x - 91 = 0$
• M1 Correct substitution: $\dfrac{-6 \pm \sqrt{36 + 728}}{4}$
• A1 $5.41$ cm