Completing the square questions with answers

Questions

1. Question 1a(x + h)² + k, x² coefficient not 1

   Express $3x^{2} + 15x - 7$ in the form $a(x + h)^2 + k$.

   Show the worked solution

   1. Step 1: Take out the coefficient of x² from the x terms$3x^{2} + 15x - 7 = 3(x^2 + 5x) -7$
   2. Step 2: Halve the coefficient of x inside the bracketHalf of $+5$ is $\dfrac{5}{2}$, so the bracket is $\left(x + \dfrac{5}{2}\right)$.
   3. Step 3: Square the bracket, then correct the number$\left(x + \dfrac{5}{2}\right)^2 = x^{2} + 5x + \dfrac{25}{4}$$3x^{2} + 15x - 7 = 3\Big(\left(x + \dfrac{5}{2}\right)^2 - \dfrac{25}{4}\Big) -7$
   4. Step 4: Multiply out and collect the numbers$= 3\left(x + \dfrac{5}{2}\right)^2 - \dfrac{75}{4} -7$$= 3\left(x + \dfrac{5}{2}\right)^2 - \dfrac{103}{4}$
2. Question 2a(x + h)² + k, x² coefficient not 1

   Express $2x^{2} - 12x + 9$ in the form $a(x + h)^2 + k$.

   Show the worked solution

   1. Step 1: Take out the coefficient of x² from the x terms$2x^{2} - 12x + 9 = 2(x^2 - 6x) +9$
   2. Step 2: Halve the coefficient of x inside the bracketHalf of $-6$ is $-3$, so the bracket is $(x - 3)$.
   3. Step 3: Square the bracket, then correct the number$(x - 3)^2 = x^{2} - 6x + 9$$2x^{2} - 12x + 9 = 2\Big((x - 3)^2 - 9\Big) +9$
   4. Step 4: Multiply out and collect the numbers$= 2(x - 3)^2 - 18 +9$$= 2(x - 3)^2 - 9$
3. Question 3Solve, x² coefficient not 1

   Solve $4x^{2} - 4x - 9 = 0$ by completing the square, giving your answers correct to 2 decimal places.

   Show the worked solution

   1. Step 1: Divide every term by the coefficient of x²$x^2 - x - \dfrac{9}{4} = 0$
   2. Step 2: Halve the coefficient of xHalf of $-1$ is $\dfrac{-1}{2}$, so the bracket is $\left(x - \dfrac{1}{2}\right)$.
   3. Step 3: Square the bracket, then correct the number$\left(x - \dfrac{1}{2}\right)^2 = x^{2} - x + \dfrac{1}{4}$$\left(x - \dfrac{1}{2}\right)^2 - \dfrac{1}{4} - \dfrac{9}{4} = 0$
   4. Step 4: Make the square the subject, then square-root both sides$\left(x - \dfrac{1}{2}\right)^2 = \dfrac{5}{2}$$x - \dfrac{1}{2} = \pm\sqrt{\dfrac{5}{2}}$
   5. Step 5: Solve and round$x = \dfrac{1}{2} \pm \sqrt{\dfrac{5}{2}}$$x = 2.0811\ldots$ or $x = -1.0811\ldots$$x = 2.08$ or $x = -1.08$ (2 d.p.)
4. Question 4Solve, x² coefficient not 1

   Solve $3x^{2} + 18x + 10 = 0$ by completing the square, giving your answers correct to 2 decimal places.

   Show the worked solution

   1. Step 1: Divide every term by the coefficient of x²$x^2 + 6x + \dfrac{10}{3} = 0$
   2. Step 2: Halve the coefficient of xHalf of $+6$ is $3$, so the bracket is $(x + 3)$.
   3. Step 3: Square the bracket, then correct the number$(x + 3)^2 = x^{2} + 6x + 9$$(x + 3)^2 - 9 + \dfrac{10}{3} = 0$
   4. Step 4: Make the square the subject, then square-root both sides$(x + 3)^2 = \dfrac{17}{3}$$x + 3 = \pm\sqrt{\dfrac{17}{3}}$
   5. Step 5: Solve and round$x = -3 \pm \sqrt{\dfrac{17}{3}}$$x = -0.6195\ldots$ or $x = -5.3804\ldots$$x = -0.62$ or $x = -5.38$ (2 d.p.)