2.5 Finding the equation | Equations and inequalities

2.5 Finding the equation (EMBFN)

We have seen that the roots are the solutions obtained from solving a quadratic equation. Given the roots, we are also able to work backwards to determine the original quadratic equation.

Worked example 11: Finding an equation when the roots are given

Find an equation with roots \(\text{13}\) and \(-\text{5}\).

Assign a variable and write roots as two equations

\[x = 13 \text{ or } x = -5\]

Use additive inverses to get zero on the right-hand sides \[x - 13 = 0 \text{ or } x + 5 = 0\]

Write down as the product of two factors

\[(x-13)(x+5) = 0\]

Notice that the signs in the brackets are opposite of the given roots.

Expand the brackets

\[x^2 - 8x - 65 = 0\]

Note that if each term in the equation is multiplied by a constant then there could be other possible equations which would have the same roots. For example,

Multiply by \(\text{2}\): \[2x^2 - 16x - 130 = 0\]

Multiply by \(-\text{3}\): \[-3x^2 + 24x + 195 = 0\]

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Worked example 12: Finding an equation when the roots are fractions

Find an equation with roots \(-\frac{3}{2}\) and \(\text{4}\).

Assign a variable and write roots as two equations

\[x = 4 \text{ or } x = -\frac{3}{2}\]

Use additive inverses to get zero on the right-hand sides. \[x - 4 = 0 \text{ or } x + \frac{3}{2} = 0\]

Multiply the second equation through by \(\text{2}\) to remove the fraction. \[x - 4 = 0 \text{ or } 2x + 3 = 0\]

Write down as the product of two factors

\[(2x+3)(x-4) = 0\]

Expand the brackets

The quadratic equation is \(2x^2 - 5x - 12 = 0\).

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Finding the equation

Textbook Exercise 2.5

Determine a quadratic equation for a graph that has roots \(\text{3}\) and \(-\text{2}\).

\[x = 3 \text{ or } x = -2\]

Use additive inverses to get zero on the right-hand sides: \[x - 3 = 0 \text{ or } x + 2 = 0\]

Write down as the product of two factors

\[(x - 3)(x + 2) = 0\]

Expand the brackets

\[x^2 - x - 6 = 0\]

Find a quadratic equation for a graph that has \(x\)-intercepts of \((-4;0)\) and \((4;0)\).

The \(x\)-intercepts are the same as the roots of the equation, so:

\[x = -4 \text{ or } x = 4\]

Use additive inverses to get zero on the right-hand sides: \[x + 4 = 0 \text{ or } x - 4 = 0\]

Write down as the product of two factors

\[(x + 4)(x - 4) = 0\]

Expand the brackets

\[x^2 - 16 = 0\]

Determine a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a, b\) and \(c\) are integers, that has roots \(-\frac{1}{2}\) and \(\text{3}\).

\[x = -\frac{1}{2} \text{ or } x = 3\]

Use additive inverses to get zero on the right-hand sides: \[x + \frac{1}{2} = 0 \text{ or } x - 3 = 0\]

Write down as the product of two factors

\[(2x + 1)(x - 3) = 0\]

Expand the brackets

\[2x^2 - 5x - 3 = 0\]

We note that this solution gives us integer values as required.

Determine the value of \(k\) and the other root of the quadratic equation \(kx^2 - 7x + 4 = 0\) given that one of the roots is \(x=1\).

We use the given root to find \(k\):

\begin{align*} k(1)^{2} - 7(1) + 4 & = 0 \\ k & = 3 \end{align*}

So the equation is:

\(3x^2 - 7x + 4 = 0\)

Now we can find the other root:

\begin{align*} 3x^{2} - 7x + 4 & = 0 \\ (x - 1)(3x - 4) & = 0 \\ x = 1 & \text{ or } x = \frac{3}{4} \end{align*}

One root of the equation \(2x^2 - 3x = p\) is \(2\frac{1}{2}\). Find \(p\) and the other root.

We use the given root to find \(p\):

\begin{align*} 2 \left( \frac{5}{2} \right)^{2} - 3 \left( \frac{5}{2} \right) - p & = 0 \\ \frac{25}{2} - \frac{15}{2} & = p \\ p & = 5 \end{align*}

So the equation is:

\(2x^2 - 3x - 5 = 0\)

Now we can find the other root:

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

Mixed exercises

Textbook Exercise 2.6

Solve the following quadratic equations by either factorisation, using the quadratic formula or completing the square:

Always try to factorise first, then use the formula if the trinomial cannot be factorised.
In a test or examination, only use the method of completing the square when specifically asked.
Answers can be left in surd or decimal form.

\(24{y}^{2}+61y-8=0\)

\begin{align*} 24y^{2} + 61y - 8 & = 0 \\ (8y - 1)(3y + 8) & = 0 \\ y = \frac{1}{8} & \text{ or } y = -\frac{8}{3} \end{align*}

\(8x^2 + 16x = 42\)

\begin{align*} 8x^{2} + 16x - 42 & = 0 \\ 4x^{2} + 8x - 21 & = 0 \\ (2x - 3)(2x + 7) & = 0 \\ x = \frac{3}{2} & \text{ or } x = -\frac{7}{2} \end{align*}

\(9t^2 = 24t - 12\)

\begin{align*} 9t^{2} - 24t + 12 & = 0 \\ 3t^{2} - 8t + 4 & = 0 \\ (3t - 2)(t - 2) & = 0 \\ t = \frac{2}{3} & \text{ or } t = 2 \end{align*}

\(-5{y}^{2}+0y+5=0\)

\begin{align*} -5y^{2} + 5 & = 0 \\ -5(y^{2} - 1) & = 0 \\ y^{2} -1 & = 0 \\ (y-1)(y+1) & = 0 \\ y = 1 & \text{ or } y = -1 \end{align*}

\(3m^2 + 12 = 15m\)

We will solve this equation by completing the square for practice:

\begin{align*} 3m^{2} + 12 - 15m & = 0 \\ m^{2} - 5m + 4 & = 0 \\ m^{2} - 5m & = -4 \\ m^{2} - 5m + \frac{25}{4} & = -4 + \frac{25}{4} \\ \left(m - \frac{5}{2}\right)^{2} & = \frac{9}{4} \\ m - \frac{5}{2} & = \pm \sqrt{\frac{9}{4}} \\ m & = \frac{5}{2} \pm \frac{3}{2} \\ m = 1 & \text{ or } m = 4 \end{align*}

\(49{y}^{2}+0y-25=0\)

\begin{align*} 49y^{2} - 25 & = 0 \\ (7y - 5)(7y + 5) & = 0 \\ y = \frac{5}{7} & \text{ or } y = -\frac{5}{7} \end{align*}

\(72 = 66w - 12w^2\)

\begin{align*} 12w^{2} - 66w + 72 & = 0 \\ 2w^{2} - 11w + 12 & = 0 \\ (2w - 3)(w - 4) & = 0 \\ w = \frac{3}{2} & \text{ or } w = 4 \end{align*}

\(-40{y}^{2}+58y-12=0\)

\begin{align*} -40y^{2} + 58y - 12 & = 0 \\ 20y^{2} - 29y + 6 & = 0 \\ (5y - 6)(4y - 1) & = 0 \\ y = \frac{6}{5} & \text{ or } y = \frac{1}{4} \end{align*}

\(37n + 72 - 24n^2 = 0\)

\begin{align*} 24n^2 - 37n - 72 & = 0 \\ (3n - 8)(8n + 9) & = 0 \\ n = \frac{8}{3} & \text{ or } n = -\frac{9}{8} \end{align*}

\(6{y}^{2}+7y-24=0\)

\begin{align*} 6{y}^{2}+7y-24 & = 0 \\ (3y + 8)(2y - 3) & = 0 \\ y = -\frac{8}{3} & \text{ or } y = \frac{3}{2} \end{align*}

\(3 = x(2x - 5)\)

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

\(-18{y}^{2}-55y-25=0\)

\begin{align*} -18y^2 - 55y - 25 &= 0\\ 18y^2 +55y + 25 &= 0 \\ y &= \dfrac{-55 \pm \sqrt{(55)^2 - (4)(18)(25)}}{2(18)}\\ &= \dfrac{-55 \pm \sqrt{\text{3 025}- \text{1 800}}}{36}\\ &= \dfrac{-55 \pm \sqrt{\text{1 225}}}{36}\\ &= \dfrac{-55 \pm 35}{36}\\ y= -\frac{90}{36} = -\frac{5}{2} &\text{ or } y=-\frac{20}{36} = -\frac{5}{9} \end{align*}

\(-25{y}^{2}+25y-4=0\)

\begin{align*} -25y^2 + 25y - 4 &= 0\\ 25y^2 - 25y+4 &= 0\\ y^2 - y &= -\frac{4}{25}\\ y^2 - y + \frac{1}{4} &= -\frac{4}{25} + \frac{1}{4}\\ \left(y-\frac{1}{2}\right)^2 &= \frac{-16+25}{100}\\ \left(y-\frac{1}{2}\right)^2 &= \frac{9}{100}\\ y-\frac{1}{2} &= \pm \sqrt{\frac{9}{100}}\\ y-\frac{1}{2} &= \pm \frac{3}{10}\\ y = \frac{1}{2}+\frac{3}{10} = \frac{4}{5} &\text{ or } y = \frac{1}{2} - \frac{3}{10} = \frac{1}{5} \end{align*}

\(8(1 - 4g^2) + 24g = 0\)

\begin{align*} -32g^2 + 24g + 8 &= 0\\ 32g^2 - 24g - 8 &= 0\\ 4g^2 - 3g - 1 &= 0\\ (4g + 1)(g-1) &= 0\\ g =-\frac{1}{4} &\text{ or } g=1 \end{align*}

\(9{y}^{2}-13y-10=0\)

\begin{align*} 9y^2 - 13y - 10&= 0\\ y &= \dfrac{-(-13) \pm \sqrt{(-13)^2 -4(9)(-10)}}{2(9)}\\ &= \dfrac{-(-13) \pm \sqrt{169 + 360}}{18}\\ &= \dfrac{13\pm \sqrt{529}}{18}\\ &= \dfrac{13\pm23}{18}\\ y = \dfrac{13+23}{18} = 2 &\text{ or } y = \dfrac{13-23}{18} = -\frac{5}{9} \end{align*}

\((7p - 3)(5p + 1) = 0\)

\begin{align*} (7p-3)(5p+1)&=0\\ p =\frac{3}{7} &\text{ or } p = -\frac{1}{5} \end{align*}

\(-81{y}^{2}-99y-18=0\)

\begin{align*} -81y^2 - 99y - 18 &= 0\\ 9y^2 + 11y +2 &=0\\ (9y+2)(y+1) &= 0\\ y = -\frac{2}{9} &\text{ or } y = -1 \end{align*}

\(14{y}^{2}-81y+81=0\)

\begin{align*} 14y^2 - 81y + 81 &= 0\\ y &= \dfrac{-(-81) \pm \sqrt{(-81)^2 - 4(14)(81)}}{2(14)}\\ &= \dfrac{81 \pm \sqrt{\text{6 561}- \text{4 536}}}{28}\\ &= \dfrac{81 \pm \sqrt{\text{2 025}}}{28}\\ &= \dfrac{81 \pm 45}{28}\\ y = \frac{81+45}{28}=\frac{9}{2} &\text{ or } y = \frac{81-45}{28} = \frac{9}{7} \end{align*}

2.4 Substitution

Table of Contents

2.6 Nature of roots