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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 ( 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

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
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
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Kristine 2*2*2=8
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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Azam
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Prasenjit
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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.
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Source:  OpenStax, Siyavula textbooks: grade 11 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11243/1.3
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