5 5 equations that may be reduced to quadratics

Equations That May be Reduced to Quadratics
Frank Ma © 2011

In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method

In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method and fractional equations that reduce to 2nd degree equations

Substitution Method
If a pattern is repeated many times in an expression, we may substitute the pattern with a variable to make the equation look simpler.

Substitution Method
Example A.
a. Given
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1

Substitution Method
Example A.
a. Given
if we substitute y for
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1
x
x – 1

Substitution Method
Example A.
a. Given
then the expression is y2 – y – 6.
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1
x
x – 1

Substitution Method
Example A.
a. Given
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1
x
x – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2

Substitution Method
Example A.
a. Given
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1
x
x – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2
if we substitute y for (x2 – 1)

Substitution Method
Example A.
a. Given
x
2
x
(
)
(
)
–
– 6
x – 1
x – 1
x
x – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2
if we substitute y for (x2 – 1)
then the expression is y2 – 3y + 2.

c. Given x4 – 5x2 – 14,

c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,

c. Given x4 – 5x2 – 14,
then the expression is y2 – 5y – 14.

c. Given x4 – 5x2 – 14,
To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations.

c. Given x4 – 5x2 – 14,
To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution,

c. Given x4 – 5x2 – 14,
To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations. We first solve the 2nd degree equations obtained after the substitution, than put the answers back to the substituted pattern and solve those equations.

c. Given x4 – 5x2 – 14,
Example B. Solve x4– 5x2 – 14 = 0 for x.

c. Given x4 – 5x2 – 14,
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)

c. Given x4 – 5x2 – 14,
(y – 7)(y + 2) = 0
y = 7, y = –2

c. Given x4 – 5x2 – 14,
(y – 7)(y + 2) = 0
y = 7, y = –2
To find x, since y = x2 , so 7 = x2 and –2 = x2

c. Given x4 – 5x2 – 14,
(y – 7)(y + 2) = 0
y = 7, y = –2
(2nd equation to solve)

c. Given x4 – 5x2 – 14,
(y – 7)(y + 2) = 0
y = 7, y = –2
(2nd equation to solve) ±7 = x ±i2 = x

x
x
2
(
)
(
)
Example C. Solve for x.
–
– 6 = 0
x – 1
x – 1

x
x
2
(
)
(
)
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
–
– 6 = 0
x – 1
x – 1
x
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
–
– 6 = 0
x – 1
x – 1
x
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x
3x – 3 = x
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x
3x – 3 = x
x = 3/2
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x –2(x – 1) = x
3x – 3 = x
x = 3/2
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x –2(x – 1) = x
3x – 3 = x –2x + 2 = x
x = 3/2
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

x
x
2
(
)
(
)
(y – 3)(y + 2) = 0
y = 3, y = –2
3(x – 1) = x –2(x – 1) = x
3x – 3 = x –2x + 2 = x
x = 3/2 2/3 = x
–
– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1

Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.

We substitute y =x1/3

We substitute y =x1/3 so x2/3 = y2,

We substitute y =x1/3 so x2/3 = y2, the equation is
2y2 + y – 6 = 0 (1st equation to solve)

(2y – 3)(y + 2) = 0
y = 3/2, y = –2

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
(2nd equation to solve)

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
(2nd equation to solve) (3/2)3 = (x1/3)3

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
(2nd equation to solve) (3/2)3 = (x1/3)3
27/8 = x

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
27/8 = x –8 = x

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
27/8 = x –8 = x
Recall the steps below for solving a rational equation.

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
27/8 = x –8 = x
I. Find the LCD.

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
27/8 = x –8 = x
I. Find the LCD.
II. Multiply both sides by the LCD to get an equation without
fractions.

(2y – 3)(y + 2) = 0
y = 3/2, y = –2
27/8 = x –8 = x
I. Find the LCD.
II. Multiply both sides by the LCD to get an equation without
fractions.
III. Solve the equation and check the answers, make sure it doesn't make the denominator 0.

x + 1
3
Example E. Solve
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
(x + 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
x2 + 2x + 1 – 2x2 + 2 = 3x – 3
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
x2 + 2x + 1 – 2x2 + 2 = 3x – 3
-x2 +2x + 3 = 3x – 3
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
x2 + 2x + 1 – 2x2 + 2 = 3x – 3
-x2 +2x + 3 = 3x – 3
0 = x2 + x – 6
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
x2 + 2x + 1 – 2x2 + 2 = 3x – 3
-x2 +2x + 3 = 3x – 3
0 = x2 + x – 6
0 = (x + 3)(x – 2)
x = –3 , x = 2
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

x + 1
3
Example E. Solve
(x – 1)(x + 1) [ ]
(x + 1) (x + 1) – 2(x – 1)(x + 1) = 3(x – 1)
x2 + 2x +1 – 2(x2 – 1) = 3x – 3
x2 + 2x + 1 – 2x2 + 2 = 3x – 3
-x2 +2x + 3 = 3x – 3
0 = x2 + x – 6
0 = (x + 3)(x – 2)
x = –3 , x = 2
Both solutions are good.
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
x + 1
3
– 2
=
x – 1
x + 1

Exercise A. Solve the following equations. Find the exact and the approximate values. If the solution is not real, state so.
1. x4 – 5x2 + 4 = 0
2. x4 – 13x2 + 36 = 0
4. 3x4+ x2– 2 = 0
3. 2x4+ x2– 6 = 0
6. 3x4 – 5x2– 2 = 0
5. 2x4+ 3x2– 2 = 0
8. 3x6+ x3– 2 = 0
7. 2x6+ x3– 6 = 0
9. 2x6+ 3x3– 2 = 0
10. 3x6 – 5x3– 2 = 0
12. 3x + x1/2– 2 = 0
11. 2x + x1/2– 6 = 0
13. 2x + 3x1/2– 2 = 0
14. 3x – 5x1/2– 2 = 0
16. 3x–2/3+ x – 1/3– 2 = 0
15. 2x–2/3+ x –1/3– 6 = 0
17. 2x–2/3+ 3x – 1/3– 2 = 0
18. 3x–2/3– 5x – 1/3– 2 = 0
2x – 3
2x – 3
)
(
)
(
19. 2 2 + – 6 = 0
x + 1
x + 1
2x – 3
2x – 3
)
(
20. 2 –2 + –1– 6 = 0
)
(
x + 1
x + 1

5 5 equations that may be reduced to quadratics

by math123b on Aug 28, 2011

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5 5 equations that may be reduced to quadratics — Presentation Transcript