<< Chapter < Page | Chapter >> Page > |
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:
Factor out the $a$ to get:
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:
and
The key is the middle term on the right hand side, which is $2\times $ the first term $\times $ the second term of the left hand side. In [link] , we know that the first term is $x$ so 2 $\times $ the second term is $\frac{b}{a}$ . This means that the second term is $\frac{b}{2a}$ . So,
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:
We set $f\left(x\right)=0$ to find its roots, which yields:
Now dividing by $a$ and taking the square root of both sides gives the expression
Finally, solving for $x$ implies that
which can be further simplified to:
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 $f\left(x\right)=2{x}^{2}+3x-7$ .
The expression cannot be factorised. Therefore, the general quadratic formula must be used.
From the equation:
Always write down the formula first and then substitute the values of $a,b$ and $c$ .
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 ${x}^{2}-5x+8=0$ .
The expression cannot be factorised. Therefore, the general quadratic formula must be used.
From the equation:
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.
Solve for $t$ using the quadratic formula.
Solve the quadratic equations by either factorisation, completing the square or by using the quadratic formula:
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. $5{y}^{2}-5=19-{y}^{2}$ |
Notification Switch
Would you like to follow the 'Siyavula textbooks: grade 11 maths' conversation and receive update notifications?