Completing the square questions with answers

Questions

1. Question 1(x + h)² + k, b even

   Express $x^{2} + 8x - 1$ in the form $(x + h)^2 + k$.

   Show the worked solution

   1. Step 1: Halve the coefficient of xHalf of $8$ is $4$, so the bracket is $(x + 4)$.
   2. Step 2: Square the bracket, then correct the number$(x + 4)^2 = x^2 +8x + 16$$x^{2} + 8x - 1 = (x + 4)^2 - 16 -1$$= (x + 4)^2 - 17$
2. Question 2(x + h)² + k, b odd

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

   Show the worked solution

   1. Step 1: Halve the coefficient of xHalf of $7$ is $\dfrac{7}{2}$, so the bracket is $\left(x + \dfrac{7}{2}\right)$.
   2. Step 2: Square the bracket, then correct the number$\left(x + \dfrac{7}{2}\right)^2 = x^2 +7x + \dfrac{49}{4}$$x^{2} + 7x - 2 = \left(x + \dfrac{7}{2}\right)^2 - \dfrac{49}{4} -2$$= \left(x + \dfrac{7}{2}\right)^2 - \dfrac{57}{4}$
3. Question 3Solve by completing the square

   Solve $x^{2} - 4x + 1 = 0$ by completing the square, giving your answers correct to 2 decimal places.

   Show the worked solution

   1. Step 1: Halve the coefficient of xHalf of $-4$ is $-2$, so the bracket is $(x - 2)$.
   2. Step 2: Square the bracket, then correct the number$(x - 2)^2 = x^2 -4x + 4$$x^{2} - 4x + 1 = (x - 2)^2 - 4 +1$$= (x - 2)^2 - 3$
   3. Step 3: Make the square the subject, then square-root both sides$(x - 2)^2 = 3$$x - 2 = \pm\sqrt{3}$
   4. Step 4: Solve and round$x = 2 \pm \sqrt{3}$$x = 3.7320\ldots$ or $x = 0.2679\ldots$$x = 3.73$ or $x = 0.27$ (2 d.p.)
4. Question 4Solve, b odd

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

   Show the worked solution

   1. Step 1: Halve the coefficient of xHalf of $1$ is $\dfrac{1}{2}$, so the bracket is $\left(x + \dfrac{1}{2}\right)$.
   2. Step 2: Square the bracket, then correct the number$\left(x + \dfrac{1}{2}\right)^2 = x^2 +x + \dfrac{1}{4}$$x^{2} + x - 3 = \left(x + \dfrac{1}{2}\right)^2 - \dfrac{1}{4} -3$$= \left(x + \dfrac{1}{2}\right)^2 - \dfrac{13}{4}$
   3. Step 3: Make the square the subject, then square-root both sides$\left(x + \dfrac{1}{2}\right)^2 = \dfrac{13}{4}$$x + \dfrac{1}{2} = \pm\sqrt{\dfrac{13}{4}}$
   4. Step 4: Solve and round$x = \dfrac{-1}{2} \pm \sqrt{\dfrac{13}{4}}$$x = 1.3027\ldots$ or $x = -2.3027\ldots$$x = 1.30$ or $x = -2.30$ (2 d.p.)
5. Question 5a(x + h)² + k, x² coefficient not 1

   Express $5x^{2} + 25x - 6$ 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$5x^{2} + 25x - 6 = 5(x^2 + 5x) -6$
   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}$$5x^{2} + 25x - 6 = 5\Big(\left(x + \dfrac{5}{2}\right)^2 - \dfrac{25}{4}\Big) -6$
   4. Step 4: Multiply out and collect the numbers$= 5\left(x + \dfrac{5}{2}\right)^2 - \dfrac{125}{4} -6$$= 5\left(x + \dfrac{5}{2}\right)^2 - \dfrac{149}{4}$
6. Question 6Solve, x² coefficient not 1

   Solve $5x^{2} + 5x - 12 = 0$ 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$x^2 + x - \dfrac{12}{5} = 0$
   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}$$\left(x + \dfrac{1}{2}\right)^2 - \dfrac{1}{4} - \dfrac{12}{5} = 0$
   4. Step 4: Make the square the subject, then square-root both sides$\left(x + \dfrac{1}{2}\right)^2 = \dfrac{53}{20}$$x + \dfrac{1}{2} = \pm\sqrt{\dfrac{53}{20}}$
   5. Step 5: Solve and round$x = \dfrac{-1}{2} \pm \sqrt{\dfrac{53}{20}}$$x = 1.1278\ldots$ or $x = -2.1278\ldots$$x = 1.13$ or $x = -2.13$ (2 d.p.)
7. Question 7Negative x² coefficient: −(x + h)² + k

   Express $-3x^{2} - 12x - 7$ in the form $a(x + h)^2 + k$.

   Show the worked solution

   1. Step 1: Take out the coefficient of x² (with its minus) from the x terms$-3x^{2} - 12x - 7 = -3(x^2 + 4x) -7$
   2. Step 2: Halve the coefficient of x inside the bracketHalf of $+4$ is $2$, so the bracket is $(x + 2)$.
   3. Step 3: Square the bracket, then correct the number$(x + 2)^2 = x^{2} + 4x + 4$$-3x^{2} - 12x - 7 = -3\Big((x + 2)^2 - 4\Big) -7$
   4. Step 4: Multiply out and collect the numbers$= -3(x + 2)^2 + 12 -7$$= -3(x + 2)^2 + 5$