Solving quadratic equations 107
Problem 18. Solve x
2
+ 2x − 8 = 0 by using the
quadratic formula
Comparing x 2 + 2x − 8 = 0 with ax 2 + bx + c = 0
gives a = 1, b = 2 and c = −8.
Substituting these values into the quadratic formula
x =
−b ±
√
b 2 − 4ac
2a
gives
x =
−2 ±
2 2 − 4(1)(−8)
2(1)
=
−2 ±
√
4 + 32
2
=
−2 ±
√
36
2
=
−2 ± 6
2
=
−2 + 6
2
or
−2 − 6
2
Hence, x =
4
2
or
−8
2
, i.e. x = 2 or x = −4.
Problem 19. Solve 3x 2 − 11x − 4 = 0 by using
the quadratic formula
Comparing 3x 2 − 11x − 4 = 0 with ax 2 + bx + c = 0
gives a = 3, b = −11 and c = −4. Hence,
x =
−(−11) ±
(−11) 2 − 4(3)(−4)
2(3)
=
+11 ±
√
121 + 48
6
=
11 ±
√
169
6
=
11 ± 13
6
=
11 + 13
6
or
11 − 13
6
Hence, x =
24
6
or
−2
6
, i.e. x = 4 or x = −
1
3
Problem 20. Solve 4x 2 + 7x + 2 = 0 giving the
roots correct to 2 decimal places
Comparing 4x 2 + 7x + 2 = 0 with ax 2 + bx + c gives
a = 4, b = 7 and c = 2. Hence,
x =
−7 ±
7 2 − 4(4)(2)
2(4)
=
−7 ±
√
17
8
=
−7 ± 4.123
8
=
−7 + 4.123
8
or
−7 − 4.123
8
Hence, x = −0.36 or −1.39, correct to 2 decimal
places.
Problem 21. Use the quadratic formula to solve
x + 2
4
+
3
x − 1
= 7 correct to 4 significant figures
Multiplying throughout by 4(x − 1) gives
4(x − 1)
(x + 2)
4
+ 4(x − 1)
3
(x − 1)
= 4(x − 1)(7)
Cancelling gives (x − 1)(x + 2) + (4)(3) = 28(x − 1)
x
2
+ x − 2 + 12 = 28x − 28
Hence,
x
2
− 27x + 38 = 0
Using the quadratic formula,
x =
−(−27) ±
(−27) 2 − 4(1)(38)
2
=
27 ±
√
577
2
=
27 ± 24.0208
2
Hence,
x =
27 + 24.0208
2
= 25.5104
or
x =
27 − 24.0208
2
= 1.4896
Hence, x = 25.51 or 1.490, correct to 4 significant
figures.
Now try the following Practice Exercise
Practice Exercise 56 Solving quadratic
equations by formula (answers on page 346)
Solve the following equations by using the
quadratic formula, correct to 3 decimal places.
1. 2x 2 + 5x − 4 = 0
2. 5.76x 2 + 2.86x − 1.35 = 0
3. 2x
2
− 7x + 4 = 0
4. 4x + 5 =
3
x
5. (2x + 1) =
5
x − 3
6. 3x 2 − 5x + 1 = 0