Challenge · The Quadratic Formula

16 marks

These problems use the quadratic formula and the discriminant. Rearrange to $=0$ first, then decide whether to solve exactly (surd form) or to a given accuracy. A worked example shows how to start; hints are at the foot. A calculator is allowed for the decimal answers.

Worked example: how to start

A rectangle has length $(x+4)$ cm and width $x$ cm. Its area is $10\text{ cm}^2$. Show that $x^2+4x-10=0$ and find $x$ to 2 d.p.

Area: $x(x+4)=10$, so $x^2+4x=10$, i.e. $x^2+4x-10=0$.

$x=\dfrac{-4\pm\sqrt{16+40}}{2}=\dfrac{-4\pm\sqrt{56}}{2}$.

$x=1.74$ (taking the positive root, since $x$ is a length).

1Show that $x^2-6x+7=0$ has two irrational roots, and write them in the form $a\pm\sqrt{b}$.[3 marks]

Answer:

2The equation $x^2+kx+16=0$ has a repeated root. Work out the possible values of $k$.[3 marks]

Answer:

3A rectangle has length $(x+2)$ cm and width $x$ cm and area $11\text{ cm}^2$. Show that $x^2+2x-11=0$, then find $x$ to 2 d.p.[4 marks]

Answer:

4Solve $3x^2-6x-2=0$, giving your answers correct to 2 decimal places.[3 marks]

Answer:

5The solutions of a quadratic $x^2+bx+c=0$ are $x=\dfrac{-3\pm\sqrt{17}}{2}$. Work out $b$ and $c$.[3 marks]

Answer:

★Extension: The equation $kx^2+4x+(k-3)=0$, where $k\neq0$, has a repeated root. Work out the possible values of $k$.

Stuck? Hints (don't peek unless you need to)1. Find the discriminant, then use the formula and simplify the surd.2. A repeated root means the discriminant is $0$.3. Form the area equation, then use the formula; $x$ is a length so take the positive root.4. Here $a=3$, so divide by $2a=6$.5. Compare with the quadratic formula when $a=1$.

Solutions & mark scheme · The Quadratic Formula

Total: 16 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1Show that $x^2-6x+7=0$ has two irrational roots, and write them in the form $a\pm\sqrt{b}$.[3]

Model solution

$b^2-4ac=36-28=8>0$, and $8$ is not a perfect square, so the roots are irrational.

$x=\dfrac{6\pm\sqrt8}{2}=\dfrac{6\pm2\sqrt2}{2}=3\pm\sqrt2$.

Answer: $3\pm\sqrt2$

Marks

✔1Discriminant $=8$ ($>0$, not a perfect square)

✔1Simplifies $\sqrt8=2\sqrt2$

✔1$x=3\pm\sqrt2$

2The equation $x^2+kx+16=0$ has a repeated root. Work out the possible values of $k$.[3]

Model solution

Repeated root: $b^2-4ac=0$, so $k^2-4(1)(16)=0$.

$k^2=64$, so $k=\pm8$.

Answer: $k=\pm8$

Marks

✔1Sets $k^2-64=0$

✔1$k^2=64$

✔1$k=8$ and $k=-8$

3A rectangle has length $(x+2)$ cm and width $x$ cm and area $11\text{ cm}^2$. Show that $x^2+2x-11=0$, then find $x$ to 2 d.p.[4]

Model solution

Area: $x(x+2)=11$, so $x^2+2x-11=0$.

$x=\dfrac{-2\pm\sqrt{4+44}}{2}=\dfrac{-2\pm\sqrt{48}}{2}$.

$\sqrt{48}=6.928\ldots$, so $x=\dfrac{-2+6.928}{2}=2.46$.

Answer: $x=2.46$

Marks

✔1Forms $x^2+2x-11=0$

✔1Uses the formula

✔1Discriminant $=48$

✔1$x=2.46$

4Solve $3x^2-6x-2=0$, giving your answers correct to 2 decimal places.[3]

Model solution

$x=\dfrac{6\pm\sqrt{36+24}}{6}=\dfrac{6\pm\sqrt{60}}{6}$.

$\sqrt{60}=7.746\ldots$, so $x=2.29$ or $x=-0.29$.

Answer: $x=2.29$ or $x=-0.29$

Marks

✔1Discriminant $=60$

✔1Divides by $2a=6$

✔1$x=2.29$ and $x=-0.29$

5The solutions of a quadratic $x^2+bx+c=0$ are $x=\dfrac{-3\pm\sqrt{17}}{2}$. Work out $b$ and $c$.[3]

Model solution

Comparing with $x=\dfrac{-b\pm\sqrt{b^2-4c}}{2}$: $-b=-3$, so $b=3$.

$b^2-4c=17$, so $9-4c=17$, giving $c=-2$.

Check: $x^2+3x-2=0$ has discriminant $9+8=17$. ✓

Answer: $b=3,\ c=-2$

Marks

✔1Compares with the formula: $b=3$

✔1$9-4c=17$

✔1$c=-2$ (with a check)

★Extension

Repeated root: $b^2-4ac=0$, so $16-4k(k-3)=0$.

$16-4k^2+12k=0$, so $k^2-3k-4=0$.

$(k-4)(k+1)=0$, so $k=4$ or $k=-1$.

Check $k=4$: $4x^2+4x+1=(2x+1)^2=0$. ✓