Year 10 Quadratic Formula Worksheets with Answers

Worked examples

Year 10A quadratic formula: worked examples

Four examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

1. Example 1 · Exact answers with the quadratic formula

   Solve $x^2 + 6x + 2 = 0$. Give your answers in exact (surd) form.

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   Key idea: For $ax^2 + bx + c = 0$, the solutions are $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Write down $a$, $b$ and $c$ first, then substitute.

   1. Read off the coefficients: $a = 1$, $b = 6$, $c = 2$.
   2. Work out the part under the square root (the discriminant):
      $b^2 - 4ac = 6^2 - 4 \times 1 \times 2 = 36 - 8 = 28$
   3. Substitute into the formula:
      $x = \frac{-6 \pm \sqrt{28}}{2 \times 1} = \frac{-6 \pm \sqrt{28}}{2}$
   4. Simplify the surd: $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$, so
      $x = \frac{-6 \pm 2\sqrt{7}}{2}$
   5. Divide every term on top by $2$:
      $x = -3 \pm \sqrt{7}$

   Answer: $x = -3 + \sqrt{7}$ or $x = -3 - \sqrt{7}$

2. Example 2 · A negative b and answers to 2 decimal places

   Solve $2x^2 - 5x - 4 = 0$, correct to 2 decimal places.

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   Key idea: Keep the signs of $b$ and $c$ with them. When $b$ is negative, $-b$ is positive, and $b^2$ is always positive.

   1. Coefficients: $a = 2$, $b = -5$, $c = -4$.
   2. Discriminant:
      $b^2 - 4ac = (-5)^2 - 4 \times 2 \times (-4) = 25 + 32 = 57$
   3. Substitute:
      $x = \frac{-(-5) \pm \sqrt{57}}{2 \times 2} = \frac{5 \pm \sqrt{57}}{4}$
   4. Work out each solution separately on your calculator:
      $x = \frac{5 + \sqrt{57}}{4} \approx 3.137 x = \frac{5 - \sqrt{57}}{4} \approx -0.637$
   5. Round at the end: $x \approx 3.14$ or $x \approx -0.64$.
   6. Quick check: $2(3.137)^2 - 5(3.137) - 4 \approx 0$ ✓

   Answer: $x \approx 3.14$ or $x \approx -0.64$

3. Example 3 · Using the discriminant to count solutions

   Use the discriminant to find how many real solutions each equation has. Don't solve them.

   (a) $3x^2 - 2x + 5 = 0$

   (b) $4x^2 - 12x + 9 = 0$

   (c) $x^2 - 3x - 7 = 0$

   Show worked solution

   Key idea: The discriminant is $\Delta = b^2 - 4ac$. If $\Delta > 0$ there are two solutions, if $\Delta = 0$ there is one, and if $\Delta < 0$ there are none (you can't take the square root of a negative number). If $\Delta$ is a perfect square, the solutions are rational and the quadratic factorises.

   1. (a) $a = 3$, $b = -2$, $c = 5$:
      $\Delta = (-2)^2 - 4 \times 3 \times 5 = 4 - 60 = -56$ $\Delta < 0$, so there are no real solutions. The parabola never touches the $x$-axis.
   2. (b) $a = 4$, $b = -12$, $c = 9$:
      $\Delta = (-12)^2 - 4 \times 4 \times 9 = 144 - 144 = 0$ $\Delta = 0$, so there is one solution, $x = \dfrac{12}{8} = 1.5$. (It's a perfect square: $(2x - 3)^2 = 0$.)
   3. (c) $a = 1$, $b = -3$, $c = -7$:
      $\Delta = (-3)^2 - 4 \times 1 \times (-7) = 9 + 28 = 37$ $\Delta > 0$, so there are two solutions. $37$ isn't a perfect square, so they are irrational: use the formula, not factorising.

   Answer: (a) none, (b) one, (c) two

4. Example 4 · A worded problem: rearrange first, then reject a negative

   A rectangular garden bed is $4$ m longer than it is wide, and its area is $20$ m². Find its width, correct to 2 decimal places.

   Show worked solution

   Key idea: Make the equation equal $0$ before reading off $a$, $b$ and $c$. Then check each answer makes sense in the context.

   1. Let the width be $x$ m. The length is $x + 4$ m, so
      $x(x + 4) = 20$
   2. Expand and rearrange so one side is $0$:
      $x^2 + 4x - 20 = 0$ So $a = 1$, $b = 4$, $c = -20$.
   3. Discriminant:
      $\Delta = 4^2 - 4 \times 1 \times (-20) = 16 + 80 = 96$
   4. Substitute:
      $x = \frac{-4 \pm \sqrt{96}}{2} = \frac{-4 \pm 4\sqrt{6}}{2} = -2 \pm 2\sqrt{6}$
   5. The two solutions are $x \approx 2.899$ and $x \approx -6.899$. A width can't be negative, so reject $-6.899$.
   6. Check: $2.899 \times 6.899 \approx 20$ ✓

   Answer: The width is about $2.90$ m (and the length about $6.90$ m).

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