<< Chapter < Page Chapter >> Page >

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 ( x ) = a x 2 + b x + c .

Factor out the a to get:

f ( x ) = a ( x 2 + b a x + c a ) .

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:

( m + n ) 2 = m 2 + 2 m n + n 2

and

( m - n ) 2 = m 2 - 2 m n + 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 b a . This means that the second term is b 2 a . So,

( x + b 2 a ) 2 = x 2 + 2 b 2 a x + ( b 2 a ) 2 .

In general if you add a quantity and subtract the same quantity, nothing has changed. This means if we add and subtract b 2 a 2 from the right hand side of [link] we will get:

f ( x ) = a ( x 2 + b a x + c a ) = a x 2 + b a x + b 2 a 2 - b 2 a 2 + c a = a x + b 2 a 2 - b 2 a 2 + c a = a x + b 2 a 2 + c - b 2 4 a

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

a ( x + b 2 a ) 2 = b 2 4 a - c

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

x + b 2 a = ± b 2 4 a 2 - c a

Finally, solving for x implies that

x = - b 2 a ± b 2 4 a 2 - c a = - b 2 a ± b 2 - 4 a c 4 a 2

which can be further simplified to:

x = - b ± b 2 - 4 a c 2 a

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 - 4 a c under the square root). These solutions are also called the roots of the quadratic equation.

Find the roots of the function f ( x ) = 2 x 2 + 3 x - 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 .

    x = - b ± b 2 - 4 a c 2 a = - ( 3 ) ± ( 3 ) 2 - 4 ( 2 ) ( - 7 ) 2 ( 2 ) = - 3 ± 65 4 = - 3 ± 65 4
  4. The two roots of f ( x ) = 2 x 2 + 3 x - 7 are x = - 3 + 65 4 and - 3 - 65 4 .

Find the solutions to the quadratic equation x 2 - 5 x + 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. x = - b ± b 2 - 4 a c 2 a = - ( - 5 ) ± ( - 5 ) 2 - 4 ( 1 ) ( 8 ) 2 ( 1 ) = 5 ± - 7 2
  4. Since the expression under the square root is negative these are not real solutions ( - 7 is not a real number). Therefore there are no real solutions to the quadratic equation x 2 - 5 x + 8 = 0 . This means that the graph of the quadratic function f ( x ) = x 2 - 5 x + 8 has no x -intercepts, but that the entire graph lies above the x -axis.

Khan academy video on quadratic equations - 2

Solution by the quadratic formula

Solve for t using the quadratic formula.

  1. 3 t 2 + t - 4 = 0
  2. t 2 - 5 t + 9 = 0
  3. 2 t 2 + 6 t + 5 = 0
  4. 4 t 2 + 2 t + 2 = 0
  5. - 3 t 2 + 5 t - 8 = 0
  6. - 5 t 2 + 3 t - 3 = 0
  7. t 2 - 4 t + 2 = 0
  8. 9 t 2 - 7 t - 9 = 0
  9. 2 t 2 + 3 t + 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 + 61 y - 8 = 0 2. - 8 y 2 - 16 y + 42 = 0 3. - 9 y 2 + 24 y - 12 = 0
4. - 5 y 2 + 0 y + 5 = 0 5. - 3 y 2 + 15 y - 12 = 0 6. 49 y 2 + 0 y - 25 = 0
7. - 12 y 2 + 66 y - 72 = 0 8. - 40 y 2 + 58 y - 12 = 0 9. - 24 y 2 + 37 y + 72 = 0
10. 6 y 2 + 7 y - 24 = 0 11. 2 y 2 - 5 y - 3 = 0 12. - 18 y 2 - 55 y - 25 = 0
13. - 25 y 2 + 25 y - 4 = 0 14. - 32 y 2 + 24 y + 8 = 0 15. 9 y 2 - 13 y - 10 = 0
16. 35 y 2 - 8 y - 3 = 0 17. - 81 y 2 - 99 y - 18 = 0 18. 14 y 2 - 81 y + 81 = 0
19. - 4 y 2 - 41 y - 45 = 0 20. 16 y 2 + 20 y - 36 = 0 21. 42 y 2 + 104 y + 64 = 0
22. 9 y 2 - 76 y + 32 = 0 23. - 54 y 2 + 21 y + 3 = 0 24. 36 y 2 + 44 y + 8 = 0
25. 64 y 2 + 96 y + 36 = 0 26. 12 y 2 - 22 y - 14 = 0 27. 16 y 2 + 0 y - 81 = 0
28. 3 y 2 + 10 y - 48 = 0 29. - 4 y 2 + 8 y - 3 = 0 30. - 5 y 2 - 26 y + 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

Questions & Answers

what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
how did I we'll learn this
Noor Reply
f(x)= 2|x+5| find f(-6)
Prince Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
AMJAD
what is system testing
AMJAD
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
Other chapter Q/A we can ask
Moahammedashifali Reply

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Siyavula textbooks: grade 11 maths' conversation and receive update notifications?

Ask