Solving by factorising questions with answers

Questions

1. Question 1x² + bx + c = 0

   Solve $x^{2} - 9x + 18 = 0$.

   Show the worked solution

   1. Step 1: Factorise$(x - 6)(x - 3) = 0$
   2. Step 2: Set each factor to zero$x - 6 = 0$ or $x - 3 = 0$$x = 6$ or $x = 3$
2. Question 2ax² + bx + c = 0

   Solve $3x^{2} - x - 10 = 0$.

   Show the worked solution

   1. Step 1: Factorise$(3x + 5)(x - 2) = 0$
   2. Step 2: Set each factor to zero$3x + 5 = 0$ or $x - 2 = 0$$x = -\dfrac{5}{3}$ or $x = 2$
3. Question 3Rearrange first

   Solve $2x^{2} = -15x -28$.

   Show the worked solution

   1. Step 1: Make one side zero$2x^{2} + 15x + 28 = 0$
   2. Step 2: Factorise$(2x + 7)(x + 4) = 0$
   3. Step 3: Set each factor to zero$2x + 7 = 0$ or $x + 4 = 0$$x = -\dfrac{7}{2}$ or $x = -4$
4. Question 4Common factor x (don't divide by x)

   Solve $2x^{2} = 3x$.

   Show the worked solution

   1. Step 1: Make one side zero$2x^{2} - 3x = 0$Dividing both sides by $x$ would lose a solution, so don't.
   2. Step 2: Factorise$x(2x - 3) = 0$
   3. Step 3: Set each factor to zero$x = 0$ or $2x - 3 = 0$$x = 0$ or $x = \dfrac{3}{2}$
5. Question 5Expand the brackets first

   Solve $x(4x - 17) = 15$.

   Show the worked solution

   1. Step 1: Expand the bracket$4x^{2} - 17x = 15$
   2. Step 2: Make one side zero$4x^{2} - 17x - 15 = 0$
   3. Step 3: Factorise$(4x + 3)(x - 5) = 0$
   4. Step 4: Set each factor to zero$4x + 3 = 0$ or $x - 5 = 0$$x = -\dfrac{3}{4}$ or $x = 5$
6. Question 6Fractional: x + k/x

   Solve $x - \dfrac{12}{x} = 4$.

   Show the worked solution

   1. Step 1: Multiply every term by the denominator$x^{2} - 12 = 4x$
   2. Step 2: Make one side zero$x^{2} - 4x - 12 = 0$
   3. Step 3: Factorise$(x + 2)(x - 6) = 0$
   4. Step 4: Set each factor to zero$x + 2 = 0$ or $x - 6 = 0$$x = -2$ or $x = 6$Neither solution makes a denominator zero, so both are valid.
7. Question 7Fractional: two fractions

   Solve $\dfrac{4}{x + 3} + \dfrac{5}{x + 4} = 2$.

   Show the worked solution

   1. Step 1: Multiply every term by the LCM of the denominators$4(x + 4) + 5(x + 3) = 2(x + 3)(x + 4)$
   2. Step 2: Expand the brackets$9x + 31 = 2x^{2} + 14x + 24$
   3. Step 3: Make one side zero$2x^{2} + 5x - 7 = 0$
   4. Step 4: Factorise$(2x + 7)(x - 1) = 0$
   5. Step 5: Set each factor to zero$2x + 7 = 0$ or $x - 1 = 0$$x = -\dfrac{7}{2}$ or $x = 1$Neither solution makes a denominator zero, so both are valid.
8. Question 8One solution given: find k, then the other

   Given that $x = -1$ is a solution of the equation $3x^{2} + kx + 7 = 0$, find the other solution.

   Show the worked solution

   1. Step 1: Substitute x = -1 into the equation$3(-1)^{2} + k(-1) + 7 = 0$
   2. Step 2: Solve for k$k = 10$
   3. Step 3: Put k back and factorise: (x + 1) is one factor$3x^{2} + 10x + 7 = 0$$(x + 1)(3x + 7) = 0$
   4. Step 4: The other factor gives the other solution$x = -\dfrac{7}{3}$