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Solving quadratic equations: the quadratic formula

Module by: Free High School Science Texts Project. E-mail the author

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 . f ( x ) = a x 2 + b x + c .
(1)

Factor out the aa to get:

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

Now we need to do some detective work to figure out how to turn Equation 2 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 ( m + n ) 2 = m 2 + 2 m n + n 2
(3)

and

( m - n ) 2 = m 2 - 2 m n + n 2 ( m - n ) 2 = m 2 - 2 m n + n 2
(4)

The key is the middle term on the right hand side, which is 2×2× the first term ×× the second term of the left hand side. In Equation 2, we know that the first term is xx so 2×× the second term is baba. This means that the second term is b2ab2a. So,

( x + b 2 a ) 2 = x 2 + 2 b 2 a x + ( b 2 a ) 2 . ( x + b 2 a ) 2 = x 2 + 2 b 2 a x + ( b 2 a ) 2 .
(5)

In general if you add a quantity and subtract the same quantity, nothing has changed. This means if we add and subtract b2a2b2a2 from the right hand side of Equation 2 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 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
(6)

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

a ( x + b 2 a ) 2 = b 2 4 a - c a ( x + b 2 a ) 2 = b 2 4 a - c
(7)

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

x + b 2 a = ± b 2 4 a 2 - c a x + b 2 a = ± b 2 4 a 2 - c a
(8)

Finally, solving for xx implies that

x = - b 2 a ± b 2 4 a 2 - c a = - b 2 a ± b 2 - 4 a c 4 a 2 x = - b 2 a ± b 2 4 a 2 - c a = - b 2 a ± b 2 - 4 a c 4 a 2
(9)

which can be further simplified to:

x = - b ± b 2 - 4 a c 2 a x = - b ± b 2 - 4 a c 2 a
(10)

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 of the expression b2-4acb2-4ac under the square root). These solutions are also called the roots of the quadratic equation.

Exercise 1: Using the quadratic formula

Find the roots of the function f(x)=2x2+3x-7f(x)=2x2+3x-7.

Solution

  1. Step 1. Determine whether the equation can be factorised :

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

  2. Step 2. Identify the coefficients in the equation for use in the formula :

    From the equation:

    a = 2 a = 2
    (11)
    b = 3 b = 3
    (12)
    c = - 7 c = - 7
    (13)
  3. Step 3. Apply the quadratic formula :

    Always write down the formula first and then substitute the values of a,ba,b and cc.

    x = - b ± b 2 - 4 a c 2 a = - ( 3 ) ± ( 3 ) 2 - 4 ( 2 ) ( - 7 ) 2 ( 2 ) = - 3 ± 65 4 = - 3 ± 65 4 x = - b ± b 2 - 4 a c 2 a = - ( 3 ) ± ( 3 ) 2 - 4 ( 2 ) ( - 7 ) 2 ( 2 ) = - 3 ± 65 4 = - 3 ± 65 4
    (14)
  4. Step 4. Write the final answer :

    The two roots of f(x)=2x2+3x-7f(x)=2x2+3x-7 are x=-3+654x=-3+654 and -3-654-3-654.

Exercise 2: Using the quadratic formula but no solution

Find the solutions to the quadratic equation x2-5x+8=0x2-5x+8=0.

Solution

  1. Step 1. Determine whether the equation can be factorised :

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

  2. Step 2. Identify the coefficients in the equation for use in the formula :

    From the equation:

    a = 1 a = 1
    (15)
    b = - 5 b = - 5
    (16)
    c = 8 c = 8
    (17)
  3. Step 3. Apply the quadratic formula :
    x = - b ± b 2 - 4 a c 2 a = - ( - 5 ) ± ( - 5 ) 2 - 4 ( 1 ) ( 8 ) 2 ( 1 ) = 5 ± - 7 2 x = - b ± b 2 - 4 a c 2 a = - ( - 5 ) ± ( - 5 ) 2 - 4 ( 1 ) ( 8 ) 2 ( 1 ) = 5 ± - 7 2
    (18)
  4. Step 4. Write the final answer :

    Since the expression under the square root is negative these are not real solutions (-7-7 is not a real number). Therefore there are no real solutions to the quadratic equation x2-5x+8=0x2-5x+8=0. This means that the graph of the quadratic function f(x)=x2-5x+8f(x)=x2-5x+8 has no xx-intercepts, but that the entire graph lies above the xx-axis.

Figure 1
Khan academy video on quadratic equations - 2

Solution by the Quadratic Formula

Solve for tt using the quadratic formula.

  1. 3 t 2 + t - 4 = 0 3 t 2 + t - 4 = 0
  2. t 2 - 5 t + 9 = 0 t 2 - 5 t + 9 = 0
  3. 2 t 2 + 6 t + 5 = 0 2 t 2 + 6 t + 5 = 0
  4. 4 t 2 + 2 t + 2 = 0 4 t 2 + 2 t + 2 = 0
  5. - 3 t 2 + 5 t - 8 = 0 - 3 t 2 + 5 t - 8 = 0
  6. - 5 t 2 + 3 t - 3 = 0 - 5 t 2 + 3 t - 3 = 0
  7. t 2 - 4 t + 2 = 0 t 2 - 4 t + 2 = 0
  8. 9 t 2 - 7 t - 9 = 0 9 t 2 - 7 t - 9 = 0
  9. 2 t 2 + 3 t + 2 = 0 2 t 2 + 3 t + 2 = 0
  10. t 2 + t + 1 = 0 t 2 + t + 1 = 0

Tip:

  • 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.
Table 1
1. 24y2+61y-8=024y2+61y-8=0 2. -8y2-16y+42=0-8y2-16y+42=0 3. -9y2+24y-12=0-9y2+24y-12=0
4. -5y2+0y+5=0-5y2+0y+5=0 5. -3y2+15y-12=0-3y2+15y-12=0 6. 49y2+0y-25=049y2+0y-25=0
7. -12y2+66y-72=0-12y2+66y-72=0 8. -40y2+58y-12=0-40y2+58y-12=0 9. -24y2+37y+72=0-24y2+37y+72=0
10. 6y2+7y-24=06y2+7y-24=0 11. 2y2-5y-3=02y2-5y-3=0 12. -18y2-55y-25=0-18y2-55y-25=0
13. -25y2+25y-4=0-25y2+25y-4=0 14. -32y2+24y+8=0-32y2+24y+8=0 15. 9y2-13y-10=09y2-13y-10=0
16. 35y2-8y-3=035y2-8y-3=0 17. -81y2-99y-18=0-81y2-99y-18=0 18. 14y2-81y+81=014y2-81y+81=0
19. -4y2-41y-45=0-4y2-41y-45=0 20. 16y2+20y-36=016y2+20y-36=0 21. 42y2+104y+64=042y2+104y+64=0
22. 9y2-76y+32=09y2-76y+32=0 23. -54y2+21y+3=0-54y2+21y+3=0 24. 36y2+44y+8=036y2+44y+8=0
25. 64y2+96y+36=064y2+96y+36=0 26. 12y2-22y-14=012y2-22y-14=0 27. 16y2+0y-81=016y2+0y-81=0
28. 3y2+10y-48=03y2+10y-48=0 29. -4y2+8y-3=0-4y2+8y-3=0 30. -5y2-26y+63=0-5y2-26y+63=0
31. x2-70=11x2-70=11 32. 2x2-30=22x2-30=2 33. x2-16=2-x2x2-16=2-x2
34. 2y2-98=02y2-98=0 35. 5y2-10=1155y2-10=115 36. 5y2-5=19-y25y2-5=19-y2

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