Quadratic formula and rearranged quadratics

(a) $0.64$

1. Use the quadratic formula with $a = 2$, $b = 5$, $c = -4$.
   $$x = \frac{-5 \pm \sqrt{25 + 32}}{4} = \frac{-5 \pm \sqrt{57}}{4}$$
2. $$x = \frac{-5 + 7.5498\ldots}{4} = 0.6374\ldots \approx 0.64$$

• M1 Substituting correctly into the formula, including $b^2 - 4ac = 25 + 32$.
• A1 The correct answer, $0.64$.

(b) $-3.14$

1. $$x = \frac{-5 - 7.5498\ldots}{4} = -3.1374\ldots \approx -3.14$$

• B1 The correct answer, $-3.14$.

(a) $-2.35$ and $0.85$

1. $$\frac{-3+\sqrt{41}}{4}$$
2. $$\frac{-3-\sqrt{41}}{4}$$
3. Use the quadratic formula with a = 2, b = 3 and c = $-4.$
4. x = ($-3$ ± $\sqrt{}$(9 + 32))/4 = ($-3$ ± $\sqrt{41}$)/4.
5. The roots are approximately $-2.35078$ and 0.85078, so x = $-2.35$ or x = 0.85 to 2 decimal places.

• M1 Establishing $\frac{-3+\sqrt{41}}{4}$ or an equivalent valid method.
• M1 Establishing $\frac{-3-\sqrt{41}}{4}$ or an equivalent valid method.
• A1 Correct answer: $-2.35$ and $0.85$

(a) $b^2 - 4ac = 16 - 28 = -12 < 0$, and a negative number has no real square root.

1. In the formula, $b^2 - 4ac = 16 - 28 = -12$.
2. You cannot take the square root of a negative number, so there are no real solutions. (Also: $(x + 2)^2 + 3 \ge 3$ is never 0.)

• B1 Working out $b^2 - 4ac = -12$ (or completing the square to $(x + 2)^2 + 3$).
• C1 Explaining that the square root of a negative number is not real, so there are no solutions.

(b) $x = -2 \pm \sqrt{11}$

1. $$x = \frac{-4 \pm \sqrt{16 + 28}}{2} = \frac{-4 \pm \sqrt{44}}{2} = \frac{-4 \pm 2\sqrt{11}}{2} = -2 \pm \sqrt{11}$$

• M1 Substituting into the formula to get $\frac{-4 \pm \sqrt{44}}{2}$.
• A1 $-2 \pm \sqrt{11}$.

(a) $\frac{-2\pm\sqrt{44}}{4}$

1. Calculate the discriminant using the signed constant.
   $$2^{2}-4\times 2\times (-5)$$
2. Use both signs over the full denominator 2a.
   $$\frac{-2+\sqrt{44}}{4}$$
3. Therefore $\frac{-2\pm\sqrt{44}}{4}$.

• M1 Calculate the discriminant using the signed constant.
• M1 Use both signs over the full denominator 2a.
• A1 Correct answer: $\frac{-2\pm\sqrt{44}}{4}$

(b) $-\frac{5}{2}$

1. Multiply the conjugate numerators over the squared denominator.
   $$\frac{4-44}{16}$$
2. Therefore $-\frac{5}{2}$.

• M1 Multiply the conjugate numerators over the squared denominator.
• A1 Correct answer: $-\frac{5}{2}$

(a) $1.14$ cm

1. Form the area equation and bring all terms to one side.
   $$2x^{2}+3x-6=0$$
2. Use the quadratic formula with the positive root for a length.
   $$\frac{-3+\sqrt{57}}{4}$$
3. Therefore $1.14$ cm.

• P1 Form the area equation and bring all terms to one side.
• P1 Use the quadratic formula with the positive root for a length.
• A1 Correct answer: $1.14$ cm

(b) $5.27$ cm

1. Substitute the unrounded width into the length expression.
   $$2\times ((-3+\sqrt{57})/4)+3$$
2. Therefore $5.27$ cm.

• P1 Substitute the unrounded width into the length expression.
• A1 Correct answer: $5.27$ cm

(a) $\frac{-4\pm\sqrt{72}}{4}$

1. Calculate the discriminant using the signed constant.
   $$4^{2}-4\times 2\times (-7)$$
2. Use both signs over the full denominator 2a.
   $$\frac{-4+\sqrt{72}}{4}$$
3. Therefore $\frac{-4\pm\sqrt{72}}{4}$.

• M1 Calculate the discriminant using the signed constant.
• M1 Use both signs over the full denominator 2a.
• A1 Correct answer: $\frac{-4\pm\sqrt{72}}{4}$

(b) $-\frac{7}{2}$

1. Multiply the conjugate numerators over the squared denominator.
   $$\frac{16-72}{16}$$
2. Therefore $-\frac{7}{2}$.

• M1 Multiply the conjugate numerators over the squared denominator.
• A1 Correct answer: $-\frac{7}{2}$

(a) $2.52$ seconds

1. Set h = 0 and rearrange.
   $$5t^{2}-12t-1.5=0$$
2. Use the quadratic formula.
   $$\frac{12+\sqrt{174}}{10}$$
3. Therefore $2.52$ seconds.

• P1 Set h = 0 and rearrange.
• P1 Use the quadratic formula.
• A1 Correct answer: $2.52$ seconds

(b) t = 0.34 s and t = 2.06 s

1. Set h = 5 and use the quadratic formula.
   $$5t^{2}-12t+3.5=0$$
2. Therefore t = 0.34 s and t = 2.06 s.

• P1 Set h = 5 and use the quadratic formula.
• A1 Correct answer: t = 0.34 s and t = 2.06 s

(a) $x = \dfrac{3 \pm \sqrt{21}}{6}$

1. Multiply both sides by $x$.
   $$x + 1 = 3x^2 - 2x$$
2. Rearrange.
   $$3x^2 - 3x - 1 = 0$$
3. Use the formula.
   $$x = \frac{3 \pm \sqrt{9 + 12}}{6} = \frac{3 \pm \sqrt{21}}{6}$$

• M1 Multiplying through by $x$: $x + 1 = 3x^2 - 2x$.
• M1 Rearranging to $3x^2 - 3x - 1 = 0$.
• M1 Substituting into the formula with $b^2 - 4ac = 9 + 12 = 21$.
• A1 $\frac{3 \pm \sqrt{21}}{6}$.