# 2.3 Finding the equation

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## Finding an equation when you know its roots

We have mentioned before that the roots of a quadratic equation are the solutions or answers you get from solving the quadatic equation. Working back from the answers, will take you to an equation.

Find an equation with roots 13 and -5

1. The step before giving the solutions would be:

$\left(x-13\right)\left(x+5\right)=0$

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

2. ${x}^{2}-8x-65=0$

Of course, there would be other possibilities as well when each term on each side of the equal to sign is multiplied by a constant.

Find an equation with roots $-\frac{3}{2}$ and 4

1. Notice that if $x=-\frac{3}{2}$ then $2x+3=0$

Therefore the two brackets will be:

$\left(2x+3\right)\left(x-4\right)=0$
2. The equation is:

$2{x}^{2}-5x-12=0$

This section is not in the syllabus, but it gives one a good understanding about some of the solutions of the quadratic equations.

## What is the discriminant of a quadratic equation?

Consider a general quadratic function of the form $f\left(x\right)=a{x}^{2}+bx+c$ . The discriminant is defined as:

$\Delta ={b}^{2}-4ac.$

This is the expression under the square root in the formula for the roots of this function. We have already seen that whether the roots exist or not depends on whether this factor $\Delta$ is negative or positive.

## Real roots ( $\Delta \ge 0$ )

Consider $\Delta \ge 0$ for some quadratic function $f\left(x\right)=a{x}^{2}+bx+c$ . In this case there are solutions to the equation $f\left(x\right)=0$ given by the formula

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}=\frac{-b±\sqrt{\Delta }}{2a}$

If the expression under the square root is non-negative then the square root exists. These are the roots of the function $f\left(x\right)$ .

There various possibilities are summarised in the figure below.

## Equal roots ( $\Delta =0$ )

If $\Delta =0$ , then the roots are equal and, from the formula, these are given by

$x=-\frac{b}{2a}$

## Unequal roots ( $\Delta >0$ )

There will be 2 unequal roots if $\Delta >0$ . The roots of $f\left(x\right)$ are rational if $\Delta$ is a perfect square (a number which is the square of a rational number), since, in this case, $\sqrt{\Delta }$ is rational. Otherwise, if $\Delta$ is not a perfect square, then the roots are irrational .

## Imaginary roots ( $\Delta <0$ )

If $\Delta <0$ , then the solution to $f\left(x\right)=a{x}^{2}+bx+c=0$ contains the square root of a negative number and therefore there are no real solutions. We therefore say that the roots of $f\left(x\right)$ are imaginary (the graph of the function $f\left(x\right)$ does not intersect the $x$ -axis).

## From past papers

1. [IEB, Nov. 2001, HG] Given:     ${x}^{2}+bx-2+k\left({x}^{2}+3x+2\right)=0\phantom{\rule{2.em}{0ex}}\left(k\ne -1\right)$
1. Show that the discriminant is given by:
$\Delta ={k}^{2}+6bk+{b}^{2}+8$
2. If $b=0$ , discuss the nature of the roots of the equation.
3. If $b=2$ , find the value(s) of $k$ for which the roots are equal.
2. [IEB, Nov. 2002, HG] Show that ${k}^{2}{x}^{2}+2=kx-{x}^{2}$ has non-real roots for all real values for $k$ .
3. [IEB, Nov. 2003, HG] The equation ${x}^{2}+12x=3k{x}^{2}+2$ has real roots.
1. Find the largest integral value of $k$ .
2. Find one rational value of $k$ , for which the above equation has rational roots.
4. [IEB, Nov. 2003, HG] In the quadratic equation $p{x}^{2}+qx+r=0$ , $p$ , $q$ and $r$ are positive real numbers and form a geometric sequence. Discuss the nature of the roots.
5. [IEB, Nov. 2004, HG] Consider the equation:
$k=\frac{{x}^{2}-4}{2x-5}\phantom{\rule{2.em}{0ex}}\mathrm{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}x\ne \frac{5}{2}$
1. Find a value of $k$ for which the roots are equal.
2. Find an integer $k$ for which the roots of the equation will be rational and unequal.
6. [IEB, Nov. 2005, HG]
1. Prove that the roots of the equation ${x}^{2}-\left(a+b\right)x+ab-{p}^{2}=0$ are real for all real values of $a$ , $b$ and $p$ .
2. When will the roots of the equation be equal?
7. [IEB, Nov. 2005, HG] If $b$ and $c$ can take on only the values 1, 2 or 3, determine all pairs ( $b;\phantom{\rule{0.222222em}{0ex}}c$ ) such that ${x}^{2}+bx+c=0$ has real roots.

## End of chapter exercises

1. Solve: ${x}^{2}-x-1=0$    (Give your answer correct to two decimal places.)
2. Solve: $16\left(x+1\right)={x}^{2}\left(x+1\right)$
3. Solve: ${y}^{2}+3+\frac{12}{{y}^{2}+3}=7$    (Hint: Let ${y}^{2}+3=k$ and solve for $k$ first and use the answer to solve $y$ .)
4. Solve for $x$ : $2{x}^{4}-5{x}^{2}-12=0$
5. Solve for $x$ :
1. $x\left(x-9\right)+14=0$
2. ${x}^{2}-x=3$    (Show your answer correct to ONE decimal place.)
3. $x+2=\frac{6}{x}$    (correct to 2 decimal places)
4. $\frac{1}{x+1}+\frac{2x}{x-1}=1$
6. Solve for $x$ by completing the square: ${x}^{2}-px-4=0$
7. The equation $a{x}^{2}+bx+c=0$ has roots $x=\frac{2}{3}$ and $x=-4$ . Find one set of possible values for $a$ , $b$ and $c$ .
8. The two roots of the equation $4{x}^{2}+px-9=0$ differ by 5. Calculate the value of $p$ .
9. An equation of the form ${x}^{2}+bx+c=0$ is written on the board. Saskia and Sven copy it down incorrectly. Saskia hasa mistake in the constant term and obtains the solutions -4 and 2. Sven has a mistake in the coefficient of $x$ and obtains the solutions 1 and -15. Determine the correct equation that was on theboard.
10. Bjorn stumbled across the following formula to solve the quadratic equation $a{x}^{2}+bx+c=0$ in a foreign textbook.
$x=\frac{2c}{-b±\sqrt{{b}^{2}-4ac}}$
1. Use this formula to solve the equation:
$2{x}^{2}+x-3=0$
2. Solve the equation again, using factorisation, to see if the formula works for this equation.
3. Trying to derive this formula to prove that it always works, Bjorn got stuck along the way. His attempt his shown below:
$\begin{array}{cccc}\hfill a{x}^{2}+bx+c& =& 0\hfill & \\ \hfill a+\frac{b}{x}+\frac{c}{{x}^{2}}& =& 0\hfill & \mathrm{Divided by}\phantom{\rule{2pt}{0ex}}{\mathrm{x}}^{2}\phantom{\rule{2pt}{0ex}}\mathrm{where}\phantom{\rule{2pt}{0ex}}\mathrm{x}\ne 0\\ \hfill \frac{c}{{x}^{2}}+\frac{b}{x}+a& =& 0\hfill & \mathrm{Rearranged}\\ \hfill \frac{1}{{x}^{2}}+\frac{b}{cx}+\frac{a}{c}& =& 0\hfill & \mathrm{Divided by}\phantom{\rule{2pt}{0ex}}\mathrm{c}\phantom{\rule{2pt}{0ex}}\mathrm{where}\phantom{\rule{2pt}{0ex}}\mathrm{c}\ne 0\\ \hfill \frac{1}{{x}^{2}}+\frac{b}{cx}& =& -\frac{a}{c}\hfill & \mathrm{Subtracted}\phantom{\rule{2pt}{0ex}}\frac{\mathrm{a}}{\mathrm{c}}\phantom{\rule{2pt}{0ex}}\mathrm{from both sides}\\ \hfill \therefore \phantom{\rule{3pt}{0ex}}\frac{1}{{x}^{2}}+\frac{b}{cx}& +& ...\hfill & \mathrm{Got stuck}\end{array}$
Complete his derivation.

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