Chapter IV

Much of the study in quadratic equation consist of different solving equation, we

have equation in quadratic form, equation containing radicals and equation reducible to

quadratic equation. They have their own steps and procedures to be followed in order to

solve the given equation.

TARGET SKILLS:

At the end of this chapter, students are expected to:

• discuss solving equation on quadratic;

• determine the index and its radicals;

• interpret the solution of the original equation; and

• select appropriate method in solving quadratic equation.

Lesson 10

Equation in Quadratic FormOBJECTIVES:

At the end of this lesson, students are expected to:

identify equation in quadratic form;

select appropriate method in solving quadratic equation; and

change the equation in standard form.

Quadratic in Form

An equation is quadratic in form when it can be written in this standard form

where the same expression is inside both (  )'s.

In other words, if you have a times the square of the expression following

b plus b times that same expression not squared plus c equal to 0, you have an

equation that is quadratic in form.

If we substitute what is in the (   ) with a variable like t, then the original

equation will become a quadratic equation.

 Solving Equations that are 

Quadratic in Form

Step 1:  Write in Standard Form,  , if needed. 

 

If it is not in standard form, move any term(s) to the appropriate side

by using the addition/subtraction property of equality. 

Also, make sure that the squared term is written first left to right, the

expression not squared is second and the constant is third and it is set

equal to 0.

Step 2: Substitute a variable in for the expression that follows b in the second term.

In other words, substitute your variable for what is in the (   ) when it is

in standard form,  .

I’m going to use t for my substitution, but really you can use any

variable as long as it is not the variable that is used in the original

equation.

Step 3: Solve the quadratic equation created in step 2. 

You can use any method you want to solve the quadratic equation:

factoring, completing the square or quadratic formula.

Step 4: Find the value of the variable from the original equation.  

Keep in mind that you are finding a solution to the original equation

and that the variable you substituted in for in step 2 is not your original

variable. 

Use the substitution that was used to set up step 2 and then solve for

the original variable.

Step 5:  Check your solutions.

In some cases, you will be working with rational exponents and square

roots in your problems.  Those types of equations can cause

extraneous solutions.  Recall that an extraneous solution is one that is

a solution to an equation after doing something like raising both sides

of an equation by an even power, but is not a solution to the original

problem.

Even though not all of the quadratic in form equations can cause

extraneous solutions, it is better to be safe than sorry and just check

them all.

Example 1: Solve the equation that is quadratic in form:  .

Standard Form, 

*Rewriting original equation to show it is

quadratic in form

*Note that (y squared) squared = y to the

fourth

*When in stand. form, let t = the

expression following b.

 

Next, we need to substitute t in for y squared in the original equation. 

 

*Original equation

 

 

*Substitute t in for y squared

Note how we ended up with a quadratic equation when we did our

substitution.  From here, we need to solve the quadratic equation that

we have created.

 

Solve the quadratic equation: factoring, completing the square or

quadratic formula.

*Factor the trinomial

 

*Use Zero-Product Principle

*Set 1st factor = 0 and solve

 

 

 

 

 

*Set 2nd factor = 0 and solve  

Let's find the value(s) of y when t = -4:

*Plug in - 4 for t

*Use square root method to solve for y

*First solution

 

 

 

 

*Second solution

 

Let's find the value(s) of y when t = 1:

*Plug in 1 for t

*Use square root method to solve for y

*First solution

 

 

 

 

*Second solution  

Example 2: Solve the equation that is quadratic in form:  .

Standard Form, 

 

*Inverse of add. 3 is sub. 3

*Equation in standard form

Note how when you square x to the 1/3 power you get x to the 2/3

power, which is what you have in the first term.

* Rewriting original equation to show it

is quadratic in form

*Note that (x to the 1/3 power) squared

= x to the 2/3 power

*When in stand. form, let t = the

expression following b.

Next, we need to substitute t in for x to the 1/3 power in the original

equation. 

*Original equation

 

 

*Substitute t in for x to the 1/3 power

   

You can use any method you want to solve the quadratic equation:

factoring, completing the square or quadratic formula.

*Factor the trinomial

 

*Use Zero-Product Principle

*Set 1st factor = 0 and solve

 

 

 

 

 

*Set 2nd factor = 0 and solve

 

 

Let's find the value(s) of x when t = 3:

*Plug in 3 for t

*Solve the rational exponent equation

*Inverse of taking it to the 1/3 power is 

raising it to the 3rd power

 

Let's find the value(s) of x when t = -1:

 

*Plug in -1 for t

*Solve the rational exponent equation

*Inverse of taking it to the 1/3 power is 

raising it to the 3rd power

Let's double check to see if x = 27 is a solution to the original

equation.

*Plugging in 27 for x

 

*True statement

 

Since we got a true statement, x = 27  is a solution.

Let's double check to see if x = -1 is a solution to the original equation.

 

*Plugging in -1 for x

 

*True statement

Since we got a true statement, x = -1  is a solution.

There are two solutions to this equation: x = 27 and x = -1.

Exercises:

1. a4 + 2a2 – 5 = 0

2. x2 – 3x + 2 = 0

3. s6 + 8s3 – 6 = 0

4. n2 – 6n + 10 = 0

5. g8 + 2g4 – g = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve the equation that is in quadratic form.

1. a8 + 2a4 – 8 = 0

_____________________________________________________

2. l2 + 4l2 – 6 = 0

_____________________________________________________

3. e4 – 8e2 – 3 = 0

_____________________________________________________

4. l6 – 10l – 5 = 0

_____________________________________________________

5. i10 – 8i5 – 4 = 0

_____________________________________________________

6. s6 – 5s3 – 25 = 0

_____________________________________________________

7. h2/4 + 8h1/4 – 12 = 0

_____________________________________________________

8. a6- 5a4 – 15 = 0

_____________________________________________________

9. n8 + 12n2 – 8 = 0

_____________________________________________________

10.e9 – 3n3 – 10 = 0

_____________________________________________________

11.x2/3 – 2x 1/3 = 8

_____________________________________________________

12.x3/6 – 3x1/2 = 9

_____________________________________________________

13.y2- 8y = 5

_____________________________________________________

14.y4 + 2y2 = 6

_____________________________________________________

15.x6 – 9x2 + 8 = 0

_____________________________________________________

Lesson 11

Equation Containing RadicalsOBJECTIVES:

At the end of this lesson, students are expected to:

determine the index and its radicals;

positively respond to the note to be remembered; and

perform isolation of one radical if there are two radicals in the equation.

In the radicals n√b which is read the “nth root of b,” the positive integer n is called

the index or order of the radical, and b is called its radicand. When n is 2, 2 is no longer

written, just simply write √b instead of 2√b to indicate the square root of b. thus 3√b is

read as “cube root of b”; 4√b as “4th root of b”.

Note:

In order to solve for x, you must isolate x.

In order to isolate x, you must remove it from under the radial.

If there are two radicals in the equation, isolate one of the radicals.

Then raise both sides of the equation to a power equal to the index of the

isolated radical.

Isolate the the remaining radical.

Raise both sides of the equation to a power equal to the index of the isolated

radical.

You should now have a polynomial equation. Solve it.

Remember that you did not start out with a polynomial; therefore, there may be

extraneous solutions. Therefore, you must check your answers.

Example 1:

First make a note of the fact that you cannot take the square root of a negative number.

Therefore, the term is valid only if and the second term is valid if

Isolate the term.

Square both sides of the equation.

Isolate the term.

Square both sides of the equation.

Check the solution by substituting 9 in the original equation for x. If the left side of the

equation equals the right side of the equation after the substitution, you have found the

correct answer.

Left side:

Right Side:1

Since the left side of the original equation does not equal the right side of the original

equation after we substituted our solution for x, then there is no solution.

You can also check the answer by graphing the equation:

.The graph represents the right side of the original equation minus the left side of the

original equation.. The x-intercept(s) of this graph is (are) the solution(s). Since there

are no x-intercepts, there are no solutions.

Exercises:

Solve each of the following equation.

1. √ x+1 = 3

2. √ x2−2x = x + 1

3. √5 x−1 = √11−x

4. √3 x+1 = 5

5. √5+3 x = 10

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve each of the following equation.

1. √3 x+1=5

_____________________________________________________

2. √1−2 x=3

_____________________________________________________

3. √5+3 x=10

_____________________________________________________

4. 2√ x+1=3

_____________________________________________________

5. 3√2x+1=2

_____________________________________________________

6. 3√2x=4

_____________________________________________________

7. 3√2(x+1)=1

_____________________________________________________

8. 3√ x−3=2

_____________________________________________________

9. √ x+1=√3x+2

_____________________________________________________

10.√2x−5=√ x+2

_____________________________________________________

11.√3 x−6=2√ x2 _____________________________________________________

12. 3√2x−3=3√ x+1

_____________________________________________________

13. 5√3 x−2=3√ x−5

_____________________________________________________

14.2√3 x−8=√4 x+1

_____________________________________________________

15.√ x−x+5=x−1

_____________________________________________________

Lesson 12

Equations Reducible to Quadratic Equations

OBJECTIVES:

At the end of this lesson, students are expected to:

interpret the solution of the original equation;

organize the equation if it is quadratic equation ; and

solve the equation by factoring or quadratic formula.

A variety of equations can be transformed into quadratic equations and solved by

methods that we have discussed in the previous section. We will consider fractional

equations, equations involving radicals and equation that can be transformed into

quadratic equations by appropriate substitutions. Since the transformation process may

introduce extraneous roots which are not solutions of the original equation, we must

always check the solution in the original equation.

Example: Solve 1 1 7

x+2 + x+3 = 12

Solution: First note that neither -2 nor -3 can be a solution since at either

of these points the equation is meaningless.

Multiplying by the LCD, 12(x+2) (x+3), we get

12(x+3) + 12(x+2) = 7(x+2) (x+3)

24x + 60 = 7(x2 + 5x + 6)

or

7x2 + 11x – 18 = 0

Factoring, we get, (7x + 18)(x – 1) = 0

x = 1 or -18

7

If x = 1, _1_ _1_ _1_ _1_ _7_

1+2 1+3 = 3 + 4 = 12

Therefore x = 1 is a solution.

If x = -18, __1__ + __1__

7 -18/7 + 2 -18/7 + 3

= __7__ + __7__

-18 + 14 -18 + 21

= _-7_ + _7_ = _7_

4 3 12

Therefore, x = _-18_ is a solution.

7

Example 2. √3 x+4 = √ x+16 - 2

Solution: squaring both sides of the equation, we obtain,

(√3 x+4¿¿2 = √ x+6+ 4

2x – 16 = 4 √ x+16

Dividing both sides by 2 gives, x – 8 = -2 √ x+16

Squaring both sides of the equation we get

(x - 8) = ¿2

x2 - 16x + 64 = 4 (x +16)

x2 – 20x = 0

x(x – 20) =0

x = 0 or x = 20

Check: if x = 0, √3 (0 )+4 = √20+16−2

√64 = √36−2

8 = 6 – 2

8 ≠ 4

Therefore x = 20 is not a solution of the original equation.

Thus the only root of √3 x+4=√x+16−2 is0.

Many equations are not quadratics equations. However, we can

transform them by means of appropriate substitutions into quadratics

equations and then solve these by techniques that we know.

Example: Solve:

a. 2x-2 – 7x-1 + 3 = 0

a. x4 – 2x2 – 2 = 0

b. ( x4 x+1 )2 –

x4 x+1❑

−2=0

Solutions

a. Let u = x-1. Then u2 = (x-1)2 = x-2 and our equation becomes

2u2 – 7u + 3 = 0, a quadratic equation in u.

To solve the equation, we factor the left-hand side.

(2u – 1)( u – 3) = 0

U = ½ or u = 3

Since u = x-1, x-1= 1x=12

or x-1 = 3, from which

x = 2 or x = 13

Check: if x = 2, 2(2-2) – 7(2-1) + 3 = 24−72+3=0

Thus x = 2 is solution

If x = 1/3, 2(1/3)-2 – 7(1/3)-1 + 3 = 2(3)2 – 7(3) + 3

So, x = 1/3 is a solution.

b. Let u = x2. Then u2 = x4 and the given equation becomes a

quadratic equation in u.

u2 – 2u – 2 = 0

By the quadratic formula;

u = 2±√ (−2 )2−4 (1 ) (−2 )

2 (1 ) = 2±2√1+2

2=1±√3

u = 1 + √3 or u = 1−√3

Since u = x2 and u = 1−√3 < 0, we have to discard this solution.

u = x2 = 1+√3 implies

x = ± √1+√3

It is simple to verify that both values of x satisfy the original

equation. The roots of x4 – x2 – 2 = 0 are √1+√3 and - √1+√3.

c. Let u = x

4 x+1 . This substitution yields a quadratic equation in u.

u2 – u – 2 = 0

(u – 2)(u + 1) = 0

u = 2 0r u = ˉ1

u = x

4 x+1 = 2 implies x = 2(4x + 1)

or x = ˉ 27

u = x

4 x+1 = 1 implies x = ˉ4x – 1

or x = ˉ 15

Again, it can easily be verified that both solutions check in the original equation.

The roots are ˉ 27

and ˉ 15

.

Exercise:

Solve the following equation by transforming each into a quadratic equation by a

suitable substitution.

1. y - √ y - 20 = 0

2. r6 + 26 r3 – 27 = 0

3. (x – 4)3 + (x – 4)3/2 – 6 = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve the following equation.

1.1x+1

+ 43x+6

=23

_____________________________________________________

2.1x+ 1x+1

= 920

_____________________________________________________

3.x+2x+3

=2 x−33 x−7

_____________________________________________________

4.x−1x

+ xx−1

=3

_____________________________________________________

5.3x−1215

=16−3 xx+6

+ x+35

_____________________________________________________

6. 3 ( x+2 )=2 (6−x 2 )x

_____________________________________________________

7.x−2 x−3

( x−1 )(x−2)+ 3 x+1x+1

=5

_____________________________________________________

8. √ x−2+1=√x+1

_____________________________________________________

9. √ x+4=2 x+7

_____________________________________________________

10.√ x+6=2 x+6

_____________________________________________________

11.√ x−3 x+4=2 x+2

_____________________________________________________

12.√3 x+4=√x+16−3

_____________________________________________________

13.√3 x+2+√6 x=4

_____________________________________________________

14.√5 y+19+ y=7

_____________________________________________________

15.√ x+3−√ x−2= 1

√ x+3

_____________________________________________________

A. Solve for x.

1. √ x+4 = 4

2. √3 x+4 - 5 = 0

3. √4 x = 45

4. √3 x−5 = √ x+95. √ x ²+15x = x + 2

B. Reduce to quadratic equation.

1. x4 – 5x + 4 = 0

2. 4(x + 3) + 5√ x+3 = 21

3. x2/3 – 5x1/3 – 6 = 0

4. (x2 + 4x)2 – (x2 + 4x) = 20

5. 2x4 – 9x2 + 7 = 0

C. Solve for x.

1.6x+1 + 1

1−x = 2x

2.1x−2 + 3

x−3 + 52−x = 0

3.2x+13

+ 22x+1 = 5

3

4.x−1x+2 + x+3

x−2 = 2

5.x−1x+2 + x+3x ²−4 =

3 x−15(x−2)