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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
Equations That May be Reduced to Quadratics
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.
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
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 ( xx – 1 )
2
– ( xx – 1 ) – 6
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
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
( xx – 1 )
2
– ( xx – 1 ) – 6
xx – 1
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
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.
( xx – 1 )
2
– ( xx – 1 ) – 6
xx – 1
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
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.
( xx – 1 )
2
– ( xx – 1 ) – 6
xx – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2
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
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.
( xx – 1 )
2
– ( xx – 1 ) – 6
xx – 1
b. Given (x2 – 1)2 – 3(x2 – 1) + 2
if we substitute y for (x2 – 1)
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
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.
( xx – 1 )
2
– ( xx – 1 ) – 6
xx – 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,
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.
Equations That May be Reduced to Quadratics
To solve an equation via substitution, instead of solving one
difficult equation, we solve two easy equations.
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
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,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 14.
Equations That May be Reduced to Quadratics
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,
if we substitute y = x2 so x4 = y2,
then the expression is y2 – 5y – 14.
Equations That May be Reduced to Quadratics
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.
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
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)
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
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
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
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
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
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)
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
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 ±i2 = x
Example C. Solve for x.( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
Equations That May be Reduced to Quadratics
Example C. Solve for x.
We substitute y = the equation is
y2 – y – 6 = 0 (1st equation to solve)
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
Equations That May be Reduced to Quadratics
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)
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 1
Equations That May be Reduced to Quadratics
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
( xx – 1 )
2– ( x
x – 1 ) – 6 = 0
xx – 1
xx – 1
xx – 1
xx – 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
Equations That May be Reduced to Quadratics
Recall the steps below for solving a rational equation.
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
Recall the steps below for solving a rational equation.
I. Find the LCD.
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
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.
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
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.
Example E. Solve x + 1x – 1
– 2 =3
x + 1
Equations That May be Reduced to Quadratics
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
x + 1x – 1
– 2 =3
x + 1
Equations That May be Reduced to Quadratics
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
Equations That May be Reduced to Quadratics
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1)
Equations That May be Reduced to Quadratics
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1)
Equations That May be Reduced to Quadratics
Example E. Solve
The LCD is (x – 1)(x + 1), multiply the LCD to both sides,
(x – 1)(x + 1) [ ]
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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)
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
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.
x + 1x – 1
– 2 =3
x + 1
x + 1x – 1
– 2 =3
x + 1
(x + 1) (x – 1)(x + 1) ( x – 1)
Equations That May be Reduced to Quadratics
Equations That May be Reduced to QuadraticsExercise 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
3. 2x4 + x2 – 6 = 0
2. x4 – 13x2 + 36 = 0
4. 3x4 + x2 – 2 = 0
5. 2x4 + 3x2 – 2 = 0 6. 3x4 – 5x2 – 2 = 0
7. 2x6 + x3 – 6 = 0 8. 3x6 + x3 – 2 = 0
9. 2x6 + 3x3 – 2 = 0 10. 3x6 – 5x3 – 2 = 0
11. 2x + x1/2 – 6 = 0 12. 3x + x1/2 – 2 = 0
13. 2x + 3x1/2 – 2 = 0 14. 3x – 5x1/2 – 2 = 0
15. 2x –2/3 + x –1/3 – 6 = 0 16. 3x –2/3 + x – 1/3 – 2 = 0
17. 2x –2/3 + 3x – 1/3 – 2 = 0 18. 3x –2/3 – 5x – 1/3 – 2 = 0
19. 2 2 + – 6 = 02x – 3x + 1)(2x – 3
x + 1)(
20. 2 –2 + –1 – 6 = 0x + 1 )(2x – 3 2x – 3
x + 1)(
