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# 5 5 equations that may be reduced to quadratics

## by math123b on Aug 29, 2011

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## 5 5 equations that may be reduced to quadraticsPresentation Transcript

• Equations That May be Reduced to Quadratics
• Equations That May be Reduced to Quadratics
In this section, we solve equations that may be reduced to 2nd degree equations by the substitution method
• Equations That May be Reduced to Quadratics
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
• Equations That May be Reduced to Quadratics
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.
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
x
2
x
(
)
(
)

– 6
x – 1
x – 1
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
if we substitute y for
x
2
x
(
)
(
)

– 6
x – 1
x – 1
x
x – 1
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
if we substitute y for
then the expression is y2 – y – 6.
x
2
x
(
)
(
)

– 6
x – 1
x – 1
x
x – 1
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
if we substitute y for
then the expression is y2 – y – 6.
x
2
x
(
)
(
)

– 6
x – 1
x – 1
x
x – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
if we substitute y for
then the expression is y2 – y – 6.
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)
• Equations That May be Reduced to Quadratics
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.
Example A.
a. Given
if we substitute y for
then the expression is y2 – y – 6.
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.
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 14.
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 14.
To solve an equation via substitution, instead of solving one difficult equation, we solve two easy equations.
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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,
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)
(y – 7)(y + 2) = 0
y = 7, y = –2
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)
(y – 7)(y + 2) = 0
y = 7, y = –2
To find x, since y = x2 , so 7 = x2 and –2 = x2
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)
(y – 7)(y + 2) = 0
y = 7, y = –2
To find x, since y = x2 , so 7 = x2 and –2 = x2
(2nd equation to solve)
• Equations That May be Reduced to Quadratics
c. Given x4 – 5x2 – 14,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 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.
Example B. Solve x4– 5x2 – 14 = 0 for x.
We substitute y = x2 so x4 = y2, the equation is
y2 – 5y – 14 = 0 (1st equation to solve)
(y – 7)(y + 2) = 0
y = 7, y = –2
To find x, since y = x2 , so 7 = x2 and –2 = x2
(2nd equation to solve) ±7 = x ±i2 = x
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.

– 6 = 0
x – 1
x – 1
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)

– 6 = 0
x – 1
x – 1
x
x – 1
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2

– 6 = 0
x – 1
x – 1
x
x – 1
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(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
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
3(x – 1) = x

– 6 = 0
x – 1
x – 1
x
x – 1
x
x – 1
x
x
x – 1
x – 1
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
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
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
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
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
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
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
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
• Equations That May be Reduced to Quadratics
x
x
2
(
)
(
)
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
(y – 3)(y + 2) = 0
y = 3, y = –2
To find x, since y =
so 3 = and –2 = (2nd equation to solve)
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
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
We substitute y =x1/3
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
We substitute y =x1/3 so x2/3 = y2,
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
We substitute y =x1/3 so x2/3 = y2, the equation is
2y2 + y – 6 = 0 (1st equation to solve)
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve)
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3
27/8 = x
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x –8 = x
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x –8 = x
Recall the steps below for solving a rational equation.
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x –8 = x
Recall the steps below for solving a rational equation.
I. Find the LCD.
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x –8 = x
Recall the steps below for solving a rational equation.
I. Find the LCD.
II. Multiply both sides by the LCD to get an equation without
fractions.
• Equations That May be Reduced to Quadratics
Example D. Solve 2x2/3 + x1/3 – 6 = 0 for x.
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
To find x, use y = x1/3, so 3/2 = x1/3 and -2 = x1/3
(2nd equation to solve) (3/2)3 = (x1/3)3 (–2)3 = (x1/3)3
27/8 = x –8 = x
Recall the steps below for solving a rational equation.
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.
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
– 2
=
x – 1
x + 1
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
– 2
=
x – 1
x + 1
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
x + 1
3
– 2
=
x – 1
x + 1
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
(x + 1)
x + 1
3
– 2
=
x – 1
x + 1
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
– 2
=
x – 1
x + 1
(x + 1) (x – 1)(x + 1)
x + 1
3
– 2
=
x – 1
x + 1
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
x + 1
3
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(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
• Equations That May be Reduced to Quadratics
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