(b)
Let
Standard form
Factor.or Zero-factor property
or Solve.
or Substitute for u.
or Square root property
Check that the solution set is E^12 , (^1) F.
x= x= 1
1
2
x^2 = x^2 = 1 x^2
1
4
u= u= 1
1
4
4 u- 1 = 0 u- 1 = 0
14 u- 121 u- 12 = 0
4 u^2 - 5 u+ 1 = 0
4 u^2 + 1 = 5 u u=x^2.
41 x^222 + 1 = 5 x^2 x^4 = 1 x^222
4 x^4 + 1 = 5 x^2
SECTION 9.3 Equations Quadratic in Form 517
This is a
key step.(c)
Standard formLetSince this equation cannot be solved by factoring, use the quadratic formula.
Simplify.Factor.Lowest termsor
or
The solution set E 33 + 26 , 33 - 26 Fcontains four numbers. NOW TRY
x= 33 + 26 x= 33 - 26
x^2 = 3 + 26 x^2 = 3 - 26 u=x^2
u= 3 26
u=
(^2) A 3 (^26) B
2
u= 224 = 24 # 26 = 226
6 226
2
u=
6 224
2
u= a=1,b=-6,c= 3
- 1 - 62 21 - 622 - 4112132
2112
u^2 - 6 u+ 3 = 0 u=x^2.
1 x^222 - 6 x^2 + 3 = 0 x^4 = 1 x^222
x^4 - 6 x^2 + 3 = 0
x^4 = 6 x^2 - 3
NOTE Equations like those in Examples 6(a) and (b)can be solved by factoring.
Example 6(a)equation
Factor.
Factor again.Using the zero-factor property gives the same solutions obtained in Example 6(a).
Equations that cannot be solved by factoring (as in Example 6(c)) must be solved by
substitution and the quadratic formula.
1 x+ 321 x- 321 x+ 221 x- 22 = 0
1 x^2 - 921 x^2 - 42 = 0
x^4 - 13 x^2 + 36 = 0
Findbothsquare
roots in each case.NOW TRY
EXERCISE 6
Solve each equation.
(a)
(b)x^4 + 4 = 8 x^2
x^4 - 17 x^2 + 16 = 0NOW TRY ANSWERS
- (a)
(b) E 24 + 213 , 24 - 213 F5 1, 46