5 4 equations that may be reduced to quadratics-x

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)(