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Solving Quadratic Equations

4.3 Solving quadratic equations (EMA36)

A quadratic equation is an equation where the exponent of the variable is at most \(\text{2}\). The following are examples of quadratic equations:

\begin{align*} 2{x}^{2} + 2x & = 1 \\ 3{x}^{2} + 2x - 1 & = 0 \\ 0 & = -2{x}^{2} + 4x - 2 \end{align*}

Quadratic equations differ from linear equations in that a linear equation has only one solution, while a quadratic equation has at most two solutions. There are some special situations, however, in which a quadratic equation has either one solution or no solutions.

We solve quadratic equations using factorisation. For example, in order to solve \(2{x}^{2} -x - 3 = 0\), we need to write it in its equivalent factorised form as \(\left(x + 1\right)\left(2x - 3\right) = 0\). Note that if \(a \times b = 0\) then \(a = 0\) or \(b = 0\).

The following video shows an example of solving a quadratic equation by factorisation.

Video: 2FBM

Method for solving quadratic equations (EMA37)

  1. Rewrite the equation in the required form, \(a{x}^{2} + bx + c = 0\).

  2. Divide the entire equation by any common factor of the coefficients to obtain an equation of the form \(a{x}^{2} + bx + c = 0\), where \(a\), \(b\) and \(c\) have no common factors. For example \(2{x}^{2} + 4x + 2 = 0\) can be written as \({x}^{2} + 2x + 1 = 0\).

  3. Factorise \(a{x}^{2} + bx + c = 0\) to be of the form \(\left(rx + s\right)\left(ux + v\right) = 0\).

  4. The two solutions are \(\left(rx + s\right) = 0\) or \(\left(ux + v\right) = 0\), so \(x= -\dfrac{s}{r}\) or \(x = -\dfrac{v}{u}\), respectively.

  5. Check the answer by substituting it back into the original equation.

Worked example 4: Solving quadratic equations

Solve for \(x\):

\[3{x}^{2} + 2x - 1 = 0\]

The equation is already in the required form, \(a{x}^{2} + bx + c = 0\)

Factorise

\[\left(x + 1\right)\left(3x - 1\right) = 0\]

Solve for both factors

We have

\begin{align*} x + 1 & = 0 \\ \therefore x & = -1 \end{align*}

OR

\begin{align*} 3x - 1 & = 0 \\ \therefore x & = \frac{1}{3} \end{align*}

Check both answers by substituting back into the original equation

Write the final answer

The solution to \(3{x}^{2} + 2x - 1 = 0\) is \(x = -1\) or \(x = \frac{1}{3}\).

Worked example 5: Solving quadratic equations

Find the roots:

\[0 = -2{x}^{2} + 4x - 2\]

The equation is already in the required form, \(a{x}^{2} + bx + c = 0\)

Divide the equation by common factor \(-\text{2}\)

\begin{align*} -2{x}^{2} + 4x - 2 & = 0 \\ {x}^{2} - 2x + 1 & = 0 \end{align*}

Factorise

\begin{align*} \left(x - 1\right)\left(x - 1\right) & = 0 \\ {\left(x - 1\right)}^{2} & = 0 \end{align*}

The quadratic is a perfect square

This is an example of a special situation in which there is only one solution to the quadratic equation because both factors are the same.

\begin{align*} x - 1 & = 0 \\ \therefore x & = 1 \end{align*}

Check the answer by substituting back into the original equation

Write final answer

The solution to \(0 = -2{x}^{2} + 4x - 2\) is \(x = 1\).

Exercise 4.2

Write the following in standard form

\((r + 4) (5 r - 4) = -\text{16}\)

\begin{align*} (r + 4) (5 r - 4) & = -\text{16} \\ \text{5} {r}^2 -\text{4} r \text{+20} r -\text{16} + \text{16} &= \text{0}\\ \text{5} {r}^2 -\text{4} r \text{+20} r -\text{16} + \text{16} &= \text{0}\\ 5 r^{2} + 16 r & = \text{0} \end{align*}

\((3 r - 8) (2 r - 3) = -\text{15}\)

\begin{align*} (3 r - 8) (2 r - 3) & = -\text{15} \\ \text{6} {r}^2 -\text{9} r -\text{16} r \text{+24} + \text{15} &= \text{0}\\ \text{6} {r}^2 -\text{9} r -\text{16} r \text{+24} + \text{15} &= \text{0}\\ 6 r^{2} - 25 r + 39 & = \text{0} \end{align*}

\((d + 5) (2 d + 5) = \text{8}\)

\begin{align*} (d + 5) (2 d + 5) & = \text{8} \\ \text{2} {d}^2 \text{+5} d \text{+10} d \text{+25} - \text{8} &= \text{0}\\ \text{2} {d}^2 \text{+5} d \text{+10} d \text{+25} - \text{8} &= \text{0}\\ 2 d^{2} + 15 d + 17 & = \text{0} \end{align*}

Solve the following equations:

\(x^{2} +2x -15 = 0\)

\begin{align*} x^{2} +2x -15 &= 0 \\ (x -3)(x +5) &=0 \\ \therefore x = -5 & \text{ or } x = 3 \end{align*}

\(p^{2} -7p -18 = 0\)

\begin{align*} p^{2} -7p -18 &= 0 \\ (p -9)(p +2) =&0 \\ \therefore p = -2 &\text{ or } p = 9 \end{align*}

\(9x^{2} - 6x - 8 = 0\)

\begin{align*} 9x^{2} - 6x - 8 & = 0 \\ (3x + 2)(3x - 4) & = 0 \\ 3x + 2 & = 0 \\ x & = -\frac{2}{3} \\ \text{or} & \\ 3x - 4 & = 0 \\ x & = \frac{4}{3} \\ \therefore x = -\frac{2}{3} & \text{ or } x = \frac{4}{3} \end{align*}

\(5x^{2} + 21x - 54 = 0\)

\begin{align*} 5x^{2} + 21x - 54 & = 0 \\ (5x - 9)(x + 6) & = 0 \\ 5x - 9 & = 0 \\ x & = \frac{9}{5} \\ \text{or} & \\ x + 6 & = 0 \\ x & = -6 \\ \therefore x = \frac{9}{5} & \text{ or } x = -6 \end{align*}

\(4 z^{2} + 12 z + 8 = 0\)

\begin{align*} 4 z^{2} + 12 z + 8 &= 0\\ z^{2} + 3 z + 2 &= 0 \\ \left(z + 1\right) \left(z + 2\right)&= 0 \\ z = -2 &\text{ or } z = -1 \end{align*}

\(- b^{2} + 7 b - 12 = 0\)

\begin{align*} - b^{2} + 7 b - 12 &= 0 \\ b^{2} - 7 b + 12 &= 0 \\ \left(b - 4\right) \left(b - 3\right) &= 0 \\ b = 3 &\text{ or } b = 4 \end{align*}

\(- 3 a^{2} + 27 a - 54 = 0\)

\begin{align*} - 3 a^{2} + 27 a - 54 &= 0 \\ a^{2} - 9 a + 18 &= 0 \\ \left(a - 6\right) \left(a - 3\right) &= 0 \\ a = 3 &\text{ or } a = 6. \end{align*}

\(4y^{2} - 9 = 0\)

\begin{align*} 4y^{2} - 9 & = 0 \\ (2y - 3)(2y + 3) & = 0 \\ 2y - 3 & = 0 \\ y & = \frac{3}{2} \\ \text{or} & \\ 2y + 3 & = 0 \\ y & = -\frac{3}{2} \\ \therefore y = \frac{3}{2} & \text{ or } y = -\frac{3}{2} \end{align*}

\(4x^{2} + 16x - 9 = 0\)

\begin{align*} 4x^{2} + 16x - 9 & = 0 \\ (2x - 1)(2x + 9) & = 0 \\ 2x - 1 & = 0 \\ x & = \frac{1}{2} \\ \text{or} & \\ 2x + 9 & = 0 \\ y & = -\frac{9}{2} \\ \therefore x = \frac{1}{2} & \text{ or } x = -\frac{9}{2} \end{align*}

\(4x^{2} - 12x = -9\)

\begin{align*} 4x^{2} - 12x & = -9 \\ 4x^{2} - 12x + 9 & = 0 \\ (2x - 3)(2x - 3) & = 0 \\ 2x - 3 & = 0 \\ x & = \frac{3}{2} \end{align*}

\(20m + 25m^{2} = 0\)

\begin{align*} 20m + 25m^{2} & = 0 \\ 5m(4 + 5m) & = 0 \\ 5m & = 0 \\ m & = 0 \\ \text{ or } & \\ 4 + 5m & = 0 \\ m & = -\frac{4}{5} \\ \therefore m = 0 & \text{ or } m = -\frac{4}{5} \end{align*}

\(2x^{2} - 5x - 12 = 0\)

\begin{align*} 2x^{2} - 5x - 12 & = 0 \\ (2x + 3)(x - 4) & = 0 \\ 2x + 3 & = 0 \\ x & = -\frac{3}{2} \\ \text{or} & \\ x - 4 & = 0 \\ x & = 4 \\ \therefore x = -\frac{3}{2} & \text{ or } x = 4 \end{align*}

\(-75x^{2} + 290x = 240\)

\begin{align*} -75x^{2} + 290x & = 240 \\ -75x^{2} + 290x - 240 & = 0 \\ -15x^{2} + 58x - 48 & = 0 \\ (5x - 6)(3x - 8) & = 0 \\ 5x - 6 & = 0 \\ x & = \frac{6}{5} \\ \text{or} & \\ 3x - 8 & = 0 \\ x & = \frac{8}{3} \\ \therefore x = \frac{6}{5} & \text{ or } x = \frac{8}{3} \end{align*}

\(2x = \frac{1}{3}x^{2} - 3x + 14\frac{2}{3}\)

\begin{align*} 2x & = \frac{1}{3}x^{2} - 3x + 14\frac{2}{3} \\ 6x & = x^{2} - 9x + 44 \\ x^{2} - 15x + 44 & = 0 \\ (x - 4)(x - 11) & = 0 \\ x - 4 & = 0 \\ x & = 4 \\ \text{or} & \\ x - 11 & = 0 \\ x & = 11 \\ \therefore x = 4 & \text{ or } x = 11 \end{align*}

\(x^{2} - 4x = -4\)

\begin{align*} x^{2} - 4x & = -4 \\ x^{2} - 4x + 4 & = 0 \\ (x - 2)(x - 2) & = 0 \\ x - 2 & = 0 \\ x & = 2 \end{align*}

\(-x^{2} + 4x - 6 = 4x^{2} - 14x + 3\)

\begin{align*} -x^{2} + 4x - 6 & = 4x^{2} - 14x + 3 \\ 5x^{2} - 18x + 9 & = 0 \\ (5x - 3)(x - 3) & = 0 \\ 5x - 3 & = 0 \\ x & = \frac{3}{5} \\ \text{or} & \\ x - 3 & = 0 \\ x & = 3 \\ \therefore x = \frac{3}{5} & \text{ or } x = 3 \end{align*}

\(t^{2} = 3t\)

\begin{align*} t^{2} & = 3t \\ t^{2} - 3t & = 0 \\ t(t - 3) & = 0 \\ t & = 0 \\ \text{or} & \\ t - 3 & = 0 \\ t & = 3 \\ \therefore t = 0 & \text{ or } t = 3 \end{align*}

\(x^{2} - 10x = -25\)

\begin{align*} x^{2} - 10x & = -25 \\ x^{2} - 10x + 25 & = 0 \\ (x - 5)(x - 5) & = 0 \\ x - 5 & = 0 \\ x & = 5 \end{align*}

\(x^{2} = 18\)

\begin{align*} x^{2} & = 18 \\ \therefore x = \sqrt{18} & \text{ or } x = -\sqrt{18} \end{align*}

\(p^{2} - 6p = 7\)

\begin{align*} p^{2} - 6p & = 7 \\ p^{2} - 6p - 7 & = 0 \\ (p - 7)(p + 1) & = 0 \\ p - 7 & = 0 \\ p & = 7 \\ \text{or} & \\ p + 1 & = 0 \\ p & = -1 \\ \therefore p = 7 & \text{ or } p = -1 \end{align*}

\(4x^{2} - 17x - 77 = 0\)

\begin{align*} 4x^{2} - 17x - 77 & = 0 \\ (4x + 11)(x - 7) & = 0 \\ 4x + 11 & = 0 \\ x & = -\frac{11}{4} \\ \text{or} & \\ x - 7 & = 0 \\ x & = 7 \\ \therefore x = -\frac{11}{4} & \text{ or } x = 7 \end{align*}

\(14x^{2} + 5x = 6\)

\begin{align*} 14x^{2} + 5x & = 6 \\ 14x^{2} + 5x - 6 & = 0 \\ (7x + 6)(2x - 1) & = 0 \\ 7x + 6 & = 0 \\ x & = -\frac{6}{7} \\ \text{or} & \\ 2x - 1 & = 0 \\ x & = \frac{1}{2} \\ \therefore x = -\frac{6}{7} & \text{ or } x = \frac{1}{2} \end{align*}

\(2x^{2} - 2x = 12\)

\begin{align*} 2x^{2} - 2x & = 12 \\ x^{2} - x - 6 & = 0 \\ (x - 3)(x + 2) & = 0 \\ x - 3 & = 0 \\ x & = 3 \\ \text{or} & \\ x + 2 & = 0 \\ x & = -2 \\ \therefore x = 3 & \text{ or } x = -2 \end{align*}
\((2a - 3)^2 - 16 = 0\)
\begin{align*} (2a - 3)^2 - 16 &= 0 \\ (2a - 3 + 4)(2a - 3 - 4) &= 0 \\ (2a + 1)(2a - 7)&= 0 \\ \therefore a = -\frac{1}{2} &\text{ or } a = \text{3,5} \end{align*}
\((x-6)^2 - 24 = 1\)
\begin{align*} (x-6)^2 - 24 &= 1 \\ (x-6)^2 - 25 &= 0 \\ (x- 6 - 5)(x- 6 + 5) &= 0\\ (x-11)(x-1) &= 0\\ \therefore x = 11 &\text{ or } x = 1 \end{align*}

Solve the following equations (note the restrictions that apply):

\(3y = \dfrac{54}{2y}\)

Note \(y \neq 0\)

\begin{align*} 3y &= \frac{54}{2y} \\ 3y^2 &= 27 \\ y^2 &= 9 \\ y^2 - 9 &= 0 \\ (y-3)(y+3) &= 0 \\ \therefore y = 3 &\text{ or } y = -3 \end{align*}
\(\dfrac{10z}{3} = 1 - \dfrac{1}{3z}\)

Note \(z \neq 0\)

\begin{align*} \frac{10z}{3} &= 1 - \frac{1}{3z} \\ 10z^2 &= 3z - 1 \\ 10z^2 - 3z + 1 &= 0 \\ (5z + 1)(2z - 1) &= 0 \\ \therefore z = -\frac{1}{5} &\text{ or } z = \frac{1}{2} \end{align*}
\(x + 2 = \dfrac{18}{x} -1\)

Note \(x \neq 0\)

\begin{align*} x + 2 &= \frac{18}{x} -1 \\ x^2 + 2x &= 18 - x \\ x^2 + 3x - 18 &= 0 \\ (x-3)(x+6) &= 0 \\ \therefore x = 3 &\text{ or } x = -6 \end{align*}
\(y - 3 = \dfrac{5}{4} - \dfrac{1}{y}\)

Note \(y \neq 0\)

\begin{align*} y - 3 &= \frac{5}{4} - \frac{1}{y} \\ 4y^2 - 12y &= 5y - 4 \\ 4y^2 -17y + 4 &= 0\\ (4y-1)(y-4)&= 0\\ \therefore y = \frac{1}{4} &\text{ or } y = 4 \end{align*}
\(\dfrac{1}{2}(b - 1) = \dfrac{1}{3}\left(\dfrac{2}{b} + 4\right)\)

Note \(b \neq 0\)

\begin{align*} \frac{1}{2}(b - 1) &= \frac{1}{3}\left(\frac{2}{b} + 4\right) \\ 3(b - 1) &= 2\left(\frac{2}{b} + 4\right) \\ 3b - 3 &= \frac{4}{b} + 8 \\ 3b^2 - 3b &= 4 + 8b \\ 3b^2 - 11b - 4 &= 0 \\ (3b + 1)(b - 4) &= \\ \therefore b = -\frac{1}{3}b &\text{ or } b = 4 \end{align*}
\(3(y + 1) = \dfrac{4}{y} + 2\)

Note \(y \neq 0\)

\begin{align*} 3(y + 1) &= \frac{4}{y} + 2 \\ 3y + 3 &= \frac{4}{y} + 2 \\ 3y^2 + 3y &= 4 + 2y \\ 3y^2 + y - 4 &= 0 \\ (3y + 4)(y-1) &= 0 \\ \therefore y = -\frac{4}{3} &\text{ or } y = 1 \end{align*}
\((x+1)^2 - 2(x+1) - 15 = 0\)
\begin{align*} (x+1)^2 - 2(x+1) - 15 &= 0 \\ ((x+1) - 5)((x+1) + 3) &= 0\\ (x-4)(x+4) &= 0 \\ \therefore x = 4 &\text{ or } x = -4 \end{align*}
\(z^4 - 1 = 0\)
\begin{align*} z^4 - 1 &= 0 \\ (z^2 - 1)(z^2 + 1) &= 0 \\ (z- 1)(z+1)(z^2 + 1) &= 0 \\ \therefore z = 1 &\text{ or } z = -1 \end{align*}

Note that \(z^{2} + 1\) has no real solutions.

\(b^4 - 13b^2 + 36 = 0\)
\begin{align*} b^4 - 13b^2 + 36 &= 0 \\ (b^2 - 4)(b^2 - 9) &= 0 \\ (b-2)(b+2)(b-3)(b+3) &= 0 \\ \therefore b = \pm 2 &\text{ or } b = \pm 3 \end{align*}

\(\dfrac{a + 1}{3a - 4} + \dfrac{9}{2a + 5} + \dfrac{2a + 3}{2a + 5} = 0\)

\begin{align*} \frac{a + 1}{3a - 4} + \frac{9}{2a + 5} + \frac{2a + 3}{2a + 5} & = 0 \\ \frac{(a + 1)(2a + 5) + 9(3a - 4) + (2a + 3)(3a - 4)}{(3a - 4)(2a + 5)} & = 0 \\ 2a^{2} + 7a + 5 + 27a - 36 + 6a^{2} + a - 12 & = 0 \\ 8a^{2} + 35a - 43 & = 0 \\ (8a + 43)(a - 1) & = 0 \\ 8a + 43 & = 0 \\ a & = -\frac{43}{8} \\ \text{or} & \\ a - 1 & = 0 \\ a & = 1 \\ \therefore a = -\frac{43}{8} & \text{ or } a = 1 \end{align*}
\(\dfrac{x^2 - 2x - 3}{x+1} = 0\)

Note \(x \neq -1\)

\begin{align*} \frac{x^2 - 2x - 3}{x+1} &= 0 \\ \frac{(x+1)(x-3)}{x+1} &= 0 \\ \therefore x &= 3 \end{align*}
\(x + 2 = \dfrac{6x -12}{x- 2}\)

Note \(x \neq 2\)

\begin{align*} x + 2 &= \frac{6x-12}{x- 2} \\ (x+2)(x-2) &= 6x - 12 \\ x^2 - 4 &= 6x - 12\\ x^2 - 6x + 8 & = 0 \\ (x - 2)(x - 4) & = 0 \\ \therefore x &= 4 \end{align*}
\(\dfrac{3(a^2+1) +10a}{3a + 1} = 1\)

Note \(a \neq -\frac{1}{3}\)

\begin{align*} \frac{3(a^2+1)+ 10a}{3a + 1} &= 1 \\ 3(a^2+1) + 10a &= 3a + 1 \\ 3a^2 + 3 + 10a - 3a - 1 &= 0 \\ 3a^2 + 7a + 2 &= 0 \\ (3a + 1)(a + 2) &= 0 \\ \therefore a &= -2 \end{align*}

\(\dfrac{3}{9a^{2} - 3a + 1} - \dfrac{3a + 4}{27a^{3} + 1} = \dfrac{1}{9a^{2} - 1}\)

\begin{align*} \frac{3}{9a^{2} - 3a + 1} - \frac{3a + 4}{27a^{3} + 1} & = \frac{1}{9a^{2} - 1} \\ \frac{3}{9a^{2} - 3a + 1} - \frac{3a + 4}{(3a + 1)(9a^{2} - 3a + 1)} & = \frac{1}{(3a - 1)(3a + 1)} \\ \frac{3(9a^{2} - 1) - (3a - 1)(3a + 4)}{(3a + 1)(3a - 1)(9a^{2} - 3a + 1)} & = \frac{9a^{2} - 3a + 1}{(3a - 1)(3a + 1)(9a^{2} - 3a + 1)} \\ 27a^{2} - 3 - 9a^{2} - 9a + 4 & = 9a^{2} - 3a + 1 \\ 9a^{2} - 6a & = 0 \\ 3a(3a - 2) & = 0 \\ 3a & = 0 \\ a & = 0 \\ \text{or} & \\ 3a - 2 & = 0 \\ a & = \frac{2}{3} \\ \therefore a = 0 & \text{ or } a = \frac{2}{3} \end{align*}