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## Solution by the quadratic formula

It is not always possible to solve a quadratic equation by factorising and sometimes it is lengthy and tedious to solve a quadratic equation by completing the square. In these situations, you can use the quadratic formula that gives the solutions to any quadratic equation.

Consider the general form of the quadratic function:

$f\left(x\right)=a{x}^{2}+bx+c.$

Factor out the $a$ to get:

$f\left(x\right)=a\left({x}^{2}+\frac{b}{a}x+\frac{c}{a}\right).$

Now we need to do some detective work to figure out how to turn [link] into a perfect square plus some extra terms. We know that for a perfect square:

${\left(m+n\right)}^{2}={m}^{2}+2mn+{n}^{2}$

and

${\left(m-n\right)}^{2}={m}^{2}-2mn+{n}^{2}$

The key is the middle term on the right hand side, which is $2×$ the first term $×$ the second term of the left hand side. In [link] , we know that the first term is $x$ so 2 $×$ the second term is $\frac{b}{a}$ . This means that the second term is $\frac{b}{2a}$ . So,

${\left(x+\frac{b}{2a}\right)}^{2}={x}^{2}+2\frac{b}{2a}x+{\left(\frac{b}{2a}\right)}^{2}.$

In general if you add a quantity and subtract the same quantity, nothing has changed. This means if we add and subtract ${\left(\frac{b}{2a}\right)}^{2}$ from the right hand side of [link] we will get:

$\begin{array}{ccc}\hfill f\left(x\right)& =& a\left({x}^{2}+\frac{b}{a}x+\frac{c}{a}\right)\hfill \\ & =& a\left({x}^{2},+,\frac{b}{a},x,+,{\left(\frac{b}{2a}\right)}^{2},-,{\left(\frac{b}{2a}\right)}^{2},+,\frac{c}{a}\right)\hfill \\ & =& a\left({\left[x,+,\left(\frac{b}{2a}\right)\right]}^{2},-,{\left(\frac{b}{2a}\right)}^{2},+,\frac{c}{a}\right)\hfill \\ & =& a\left({\left[x,+,\left(\frac{b}{2a}\right)\right]}^{2}\right)+c-\frac{{b}^{2}}{4a}\hfill \end{array}$

We set $f\left(x\right)=0$ to find its roots, which yields:

$a{\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}}{4a}-c$

Now dividing by $a$ and taking the square root of both sides gives the expression

$x+\frac{b}{2a}=±\sqrt{\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}}$

Finally, solving for $x$ implies that

$\begin{array}{ccc}\hfill x& =& -\frac{b}{2a}±\sqrt{\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}}\hfill \\ & =& -\frac{b}{2a}±\sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \end{array}$

which can be further simplified to:

$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

These are the solutions to the quadratic equation. Notice that there are two solutions in general, but these may not always exists (depending on the sign ofthe expression ${b}^{2}-4ac$ under the square root). These solutions are also called the roots of the quadratic equation.

Find the roots of the function $f\left(x\right)=2{x}^{2}+3x-7$ .

1. The expression cannot be factorised. Therefore, the general quadratic formula must be used.

2. From the equation:

$a=2$
$b=3$
$c=-7$
3. Always write down the formula first and then substitute the values of $a,b$ and $c$ .

$\begin{array}{ccc}\hfill x& =& \frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ & =& \frac{-\left(3\right)±\sqrt{{\left(3\right)}^{2}-4\left(2\right)\left(-7\right)}}{2\left(2\right)}\hfill \\ & =& \frac{-3±\sqrt{65}}{4}\hfill \\ & =& \frac{-3±\sqrt{65}}{4}\hfill \end{array}$
4. The two roots of $f\left(x\right)=2{x}^{2}+3x-7$ are $x=\frac{-3+\sqrt{65}}{4}$ and $\frac{-3-\sqrt{65}}{4}$ .

Find the solutions to the quadratic equation ${x}^{2}-5x+8=0$ .

1. The expression cannot be factorised. Therefore, the general quadratic formula must be used.

2. From the equation:

$a=1$
$b=-5$
$c=8$
3. $\begin{array}{ccc}\hfill x& =& \frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\hfill \\ & =& \frac{-\left(-5\right)±\sqrt{{\left(-5\right)}^{2}-4\left(1\right)\left(8\right)}}{2\left(1\right)}\hfill \\ & =& \frac{5±\sqrt{-7}}{2}\hfill \end{array}$
4. Since the expression under the square root is negative these are not real solutions ( $\sqrt{-7}$ is not a real number). Therefore there are no real solutions to the quadratic equation ${x}^{2}-5x+8=0$ . This means that the graph of the quadratic function $f\left(x\right)={x}^{2}-5x+8$ has no $x$ -intercepts, but that the entire graph lies above the $x$ -axis.

## Solution by the quadratic formula

Solve for $t$ using the quadratic formula.

1. $3{t}^{2}+t-4=0$
2. ${t}^{2}-5t+9=0$
3. $2{t}^{2}+6t+5=0$
4. $4{t}^{2}+2t+2=0$
5. $-3{t}^{2}+5t-8=0$
6. $-5{t}^{2}+3t-3=0$
7. ${t}^{2}-4t+2=0$
8. $9{t}^{2}-7t-9=0$
9. $2{t}^{2}+3t+2=0$
10. ${t}^{2}+t+1=0$
• In all the examples done so far, the solutions were left in surd form. Answers can also be given in decimal form, using the calculator. Read the instructions when answering questions in a test or exam whether to leave answers in surd form, or in decimal form to an appropriate number of decimal places.
• Completing the square as a method to solve a quadratic equation is only done when specifically asked.

## Mixed exercises

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

• Always try to factorise first, then use the formula if the trinomial cannot be factorised.
• Do some of them by completing the square and then compare answers to those done using the other methods.
 1. $24{y}^{2}+61y-8=0$ 2. $-8{y}^{2}-16y+42=0$ 3. $-9{y}^{2}+24y-12=0$ 4. $-5{y}^{2}+0y+5=0$ 5. $-3{y}^{2}+15y-12=0$ 6. $49{y}^{2}+0y-25=0$ 7. $-12{y}^{2}+66y-72=0$ 8. $-40{y}^{2}+58y-12=0$ 9. $-24{y}^{2}+37y+72=0$ 10. $6{y}^{2}+7y-24=0$ 11. $2{y}^{2}-5y-3=0$ 12. $-18{y}^{2}-55y-25=0$ 13. $-25{y}^{2}+25y-4=0$ 14. $-32{y}^{2}+24y+8=0$ 15. $9{y}^{2}-13y-10=0$ 16. $35{y}^{2}-8y-3=0$ 17. $-81{y}^{2}-99y-18=0$ 18. $14{y}^{2}-81y+81=0$ 19. $-4{y}^{2}-41y-45=0$ 20. $16{y}^{2}+20y-36=0$ 21. $42{y}^{2}+104y+64=0$ 22. $9{y}^{2}-76y+32=0$ 23. $-54{y}^{2}+21y+3=0$ 24. $36{y}^{2}+44y+8=0$ 25. $64{y}^{2}+96y+36=0$ 26. $12{y}^{2}-22y-14=0$ 27. $16{y}^{2}+0y-81=0$ 28. $3{y}^{2}+10y-48=0$ 29. $-4{y}^{2}+8y-3=0$ 30. $-5{y}^{2}-26y+63=0$ 31. ${x}^{2}-70=11$ 32. $2{x}^{2}-30=2$ 33. ${x}^{2}-16=2-{x}^{2}$ 34. $2{y}^{2}-98=0$ 35. $5{y}^{2}-10=115$ 36. $5{y}^{2}-5=19-{y}^{2}$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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