Solving quadratic equations 107
Problem 18. Solvex^2 + 2 x− 8 =0byusingthe
quadratic formulaComparing x^2 + 2 x− 8 =0 with ax^2 +bx+c= 0
givesa= 1 ,b=2andc=−8.
Substituting these values into the quadratic formula
x=
−b±√
b^2 − 4 ac
2 agives
x=− 2 ±√
22 − 4 ( 1 )(− 8 )
2 ( 1 )=− 2 ±√
4 + 32
2=− 2 ±√
36
2=− 2 ± 6
2=− 2 + 6
2or− 2 − 6
2Hence,x=
4
2or− 8
2,i.e.x= 2 orx=− 4.Problem 19. Solve 3x^2 − 11 x− 4 =0byusing
the quadratic formulaComparing 3x^2 − 11 x− 4 =0 withax^2 +bx+c= 0
givesa= 3 ,b=−11 andc=−4. Hence,
x=−(− 11 )±√
(− 11 )^2 − 4 ( 3 )(− 4 )
2 ( 3 )=+ 11 ±√
121 + 48
6=11 ±√
169
6=11 ± 13
6=11 + 13
6or11 − 13
6Hence,x=
24
6or− 2
6,i.e.x= 4 orx=−1
3Problem 20. Solve 4x^2 + 7 x+ 2 =0givingthe
roots correct to 2 decimal placesComparing 4x^2 + 7 x+ 2 =0 withax^2 +bx+cgives
a= 4 ,b=7andc=2. Hence,
x=− 7 ±√
72 − 4 ( 4 )( 2 )
2 ( 4 )=− 7 ±√
17
8=− 7 ± 4. 123
8=− 7 + 4. 123
8or− 7 − 4. 123
8Hence, 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 figuresMultiplying throughout by 4(x− 1 )gives4 (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 = 28 x− 28Hence, x^2 − 27 x+ 38 = 0Using the quadratic formula,x=
−(− 27 )±√
(− 27 )^2 − 4 ( 1 )( 38 )
2=27 ±√
577
2=27 ± 24. 0208
2Hence, x=27 + 24. 0208
2= 25. 5104or x=27 − 24. 0208
2= 1. 4896Hence,x= 25. 51 or 1. 490 , correct to 4 significant
figures.Now try the following Practice ExercisePracticeExercise 56 Solving quadratic
equations by formula (answers on page 346)
Solve the following equations by using the
quadratic formula, correct to 3 decimal places.- 2x^2 + 5 x− 4 = 0
- 76 x^2 + 2. 86 x− 1. 35 = 0
- 2x^2 − 7 x+ 4 = 0
- 4x+ 5 =
3
x- ( 2 x+ 1 )=
5
x− 3- 3x^2 − 5 x+ 1 = 0