11 ALGEBRA

Solution 2		Alternatively, we can directly factorize the equation.
		x4 – 13x2 + 36 = 0
		(x2 – 4)(x2 – 9) = 0
		x2 = 4	or	x2 = 9
		x = 2		x = 3
		In both solutions, the roots of the given equation are –3, –2, 2, 3.
	2
EXAMPLE		Solve (5x – 1)2 + 4(5x – 1) – 5 = 0.

Solution

EXAMPLE 3

For the equation (5x – 1)2 + 4(5x – 1) – 5 = 0, we let t = 5x – 1 so that t2 = (5x – 1)2. Then the original equation becomes t2 + 4t – 5 = 0. First solve for t:

	(t + 5)(t – 1) = 0, so
	t = –5 or t = 1.
	Now, solve for x.
	5x – 1 = –5			or	5x – 1 = 1
	5x = –4				5x = 2
	x =	–4			x	= 2
	1	5			2	5

Hence, the roots of the equation are

Solve (a2 – a)2 – 2(a2 – a) = 0.

–4 and 2 . 5 5

Solution Let t = a2 – a.

The equation becomes t2 – 2t = 0

t(t – 2) = 0

t = 0 or t = 2. Now, solve for a.

1.a2 – a = 0 a(a – 1) = 0

a = 0 or a = 1

2.a2 – a = 2

a2 – a – 2 = 0

(a – 2)(a + 1) = 0 a = 2 or a = –1

Hence, the roots of the equation are –1, 0, 1, 2.

Equations and Inequalities

259

4

1

4

2

EXAMPLE

2

Solve x

+5 = 2x

+

.

x

x

2

Solution

First x2

0, so

x 0.

Let x

1

= t.

x

How can we write the right-hand side in terms of t?

Let’s take the square of t:

2

1 2

2

1

2, so

t

=

x

= x

+

2

x

x

x2 +

1

= t2 +2. Now, let us multiply both sides of the equation by 2:

2

x

2x2 +

2

= 2t2 +4.

2

x

Then we have the equation

t4 +5 = 2t2 +4

t4 2t2 +1= 0

(t2 1)2

0

t2 1= 0

t =1 or

t= 1.

Since t= x 1,

x

x

1

=1

or x

1= 1.

x

x

Let us multiply the equations by x:

x2 x 1= 0 or x2 + x 1= 0.

By using the quadratic formula,

x

= 1 5 , x = 1+ 5 ,

x

=

1+ 5

,

x

=

1 5

.

1

2

2

2

3

2

4

2

Check Yourself 9

Solve the equations.

1.(x + 15)2 – 3(x + 15) – 18 = 0.

2.(x + 5)4 – 6(x + 5)2 – 7 = 0.

3.x – 9ñx + 14 = 0.

Answers

1. –18, –9 2. –5 ñ7 3. 4, 49

260	Algebra 11

B. EQUATIONS INVOLVING PRODUCTS AND QUOTIENTS

If the product of two or more numbers is zero, then at least one of the factors must be zero. If the quotient of a division is zero, then the dividend must be zero.

Note

1.P Q = 0 if and only if P = 0 or Q = 0, where P = P(x) and Q = Q(x).

2.QP = 0 if and only if P = 0 and Q 0.

	5
EXAMPLE		Solve (x2 – 1)(x2 – 2x – 8) = 0.
Solution		Try to factorize each part if possible:
		(x – 1)(x + 1)(x – 4)(x + 2) = 0.
		If the product is zero then at least one of the factors is zero.
		x – 1 = 0		;	x = 1
		x + 1	= 0	;	x = –1
		x – 4	= 0	;	x = 4
		x + 2	= 0	;	x = –2
		Thus, the roots of the equation are –2, –1, 1, 4.
	6
EXAMPLE		Solve (x2 – 4x + 10)(x2 – 5x + 2) = 0.

Solution We cannot factorize the parts.

Let us try to solve x2 – 4x + 10 = 0 and x2 – 5x + 2 = 0 as two different quadratic equations.

x2 – 4x + 10 = 0; = 16 – 4 1 10 = –24 < 0. Therefore, this equation has no real root.

x2 – 5x + 2 = 0 ; = 25 – 4 1 2 = 17 This time, by the quadratic formula, the roots are

x = 5 17		, x = 5+ 17 .
1	2	2	2

Equations and Inequalities

261

EXAMPLE 7 Solve x3 – x2 – 4x + 4 = 0.

Solution First try to factorize the expression.

x3 – x2 – 4x + 4 = x2(x – 1) – 4(x – 1) = (x – 1)(x2 – 4) = (x – 1)(x – 2)(x + 2). Now, the question becomes:

	solve	(x – 1)(x – 2)(x + 2) = 0.
	So the solution is
		x – 1	= 0 ;			x1	= 1
		x – 2	= 0 ;			x2	= 2
		x + 2	= 0 ; x3				= –2.
	8 Solve	x2 5x 6			= 0.
EXAMPLE
		x+2
Solution The denominator of a fraction cannot be equal to zero, so x + 2 0,								x –2.

We need to solve x2 – 5x – 6 = 0. x2 – 5x – 6 = (x – 6)(x + 1)

(x – 6)(x + 1) = 0 x1 = –1 or x2 = 6

Since these roots are not equal to –2, x1 = –1 and x2 = 6 are both solutions to the equation.

	9	Solve	x2 +7x 8	= 0.
EXAMPLE
			x2 3x+2
Solution		First solve x2 – 3x + 2 0: x 2,			x 1.

x2 + 7x – 8 = (x – 1) (x + 8) = 0

So x1 = 1 and x2 = –8 are the roots of the numerator. We know that x 1, so the only solution is –8.

Note

It is very important to check the roots of the numerator to see whether they make the denominator zero or not. We can do this either by substituting the roots of the numerator in the denominator, or by finding the roots of the denominator directly and checking whether they are common or not.

262	Algebra 11

		10		x 7	x+3
	EXAMPLE		Solve			=1.
				x 4	x+ 4

Solution x – 4 0, and x + 4 0, so x 4, and x –4 (since the denominators cannot be zero).

Now we have	x 7	x+3	1= 0.
	x 4	x+ 4

Let us make the denominators common:

(x 7)(x+4) (x+3)(x 4) (x2 16)			= 0.
(x+4)(x 1)
	x2	2x	= 0
(x 4)(x+4)
–x2 – 2x = 0 ; –x(x + 2) = 0 ;		x1 = 0 and x2 = – 2.

		11		4		x+1	x2 5
	EXAMPLE		Solve			x 1	= x2 1 .
				x+1						x+1 x2 5
	Solution						4
										x 1 = x2 1
			In its present form, the equation x+1								is not a quadratic equation. However,

we can make it quadratic by multiplying each side by x2 – 1 since (x2 – 1 0 ; x 1). The result is 4(x – 1) – (x + 1)2 = x2 – 5

4x – 4 – x2 – 2x – 1 = x2 – 5 –2x2 + 2x = 0

–2x(x – 1) = 0

x1 = 0 or x2 = 1, but x 1, so x = 0 is the only possible solution.

	12				x	2		5x
			x+2					x+2
EXAMPLE		Solve							+6 = 0.
Solution		x + 2 0 ;				x –2
		Let t =			x		, then the equation becomes

x+2 t2 – 5t + 6 = 0

(t – 3)(t – 2) = 0 t1 = 2 or t2 = 3.

Equations and Inequalities

263

Now, solve for x:

	x	= 2	or	x	=3
	x+2			x+2
	x = 2x+ 4			x = 3x+6
	x2 = 4			x1 = 3.

So the roots are –4 and –3.

Check Yourself 10

1.	Solve		9		+2x = 4.
			4x+3
2.	Solve		x+1			x 3	=	12	.
			x 2					x2 4
						x+ 2
Answers
1.	1 ;	3	2. no real solution
	4	2

C. EQUATIONS INVOLVING RADICALS

When the variable in an equation occurs in a square root, cube root, and so on, that is, when it occurs in a radical, the equation is called a radical equation. For example, the equations x+1 = 2 and 3 x – 1 – 5 =0 are radical equations. Sometimes a suitable operation will

change a radical equation to an equation that is linear or quadratic.

To solve a radical equation, follow the procedure.

1. Isolate the radicals

Isolating a radical means putting the radical on one side of the equation and everything else on the other side, using inverse operations. If there are two radicals in the equation, isolate one of the radicals.

2. Get rid of the radical sign

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

3.If there is still a radical sign left, repeat steps 1 and 2.

4.Solve the remaining equation

264	Algebra 11

5. Check for extraneous solutions

When you solve a radical equation, extra solutions may come up when you raise both sides to an even power. These extra solutions are called extraneous solutions. In radical equations you check for extraneous solutions by putting the values you found into the original equation. If the left side of the equation does not equal the right side then you have an extraneous solution.

Note

1.If a value is an extraneous solution, it is not a solution to the original problem.

2.It is very important to check your results in the original equation. In many equations, one of the results may not satisfy the original equation. However, sometimes it is possible that all results that you have found will be acceptable.

	13			2x+5 = 7.
EXAMPLE		Solve the radical equation
Solution		Here the radicand is already alone; we do not need to isolate it. So take the square of both sides:
		2x + 5 = 49 ; 2x = 44 ;		x = 22.
		Now let us check to see if x = 22 is an extraneous solution:
		2x+5	= 7
		2 22+5	= 7
		49	= 7

7 = 7.

Since the last statement is true, x = 22 is not an extraneous solution. Therefore, there is one solution to this radical equation, x = 22.

	14	Solve 2x 5 + x= 4.
EXAMPLE
Solution		First we isolate the radical:
		2x 5 = 4 x.
		Now take the squares of both sides to eliminate the square root:
		2x – 5 = 16 – 8x + x2.
		The new equation is x2 – 10x + 21 = 0
		(x – 3) (x – 7) = 0
		x = 3 or x = 7.
		Let us check the results in the original equation:

Equations and Inequalities

265

x = 7 ; 2.7 5 +7 = 4		x = 3 ; 2 3 5 +3 = 4
9 +7	= 4	1+3	= 4
3+7	= 4	4	= 4
10	= 4	This is true, so x = 3 is a solution.

This is false, so 7 is an extraneous solution.

Hence, the only solution to the equation is x = 3.

EXAMPLE 15 Solve 10x+56 2x+8 = 4.

Solution In this question there are two radical expressions. We can isolate only one expression, so it is better to isolate the more complex one. So we have

10x+56 = 2x+8 +4. Take the squares of both sides:

10x+56 =(2x+8)+2 4 2x+8 +16.

This is a new equation involving radical expressions. Follow the same steps again to isolate the second radical.

8x+32 = 8 2x+8

x+4 = 2x+8

x2 +8x+16 = 2x+8

x2 +6x+8 = 0

(x+4) (x+2) = 0 x = –4 , x = –2

Now, check these results in the original equation.

x = –4 ; 2 (–4)+8

10 (–4) +56 = 4

0 16 = 4

4 = 4 This is true!

x = –2 ; 2 (–2)+8

10 (–2)+56 = 4

4 36 = 4

2 6 = 4

4 = 4 This is true!

Hence, both –4 and –2 are solutions to the equation.

266	Algebra 11

EXAMPLE 16 Solve the equation 5+ 3 x+3 = 3.

Solution 5+ 3 x+3 = 3

3 x+3 = 2 (by taking the cube of both sides)

x+3 = 8

x = –11. We do not need to check for extraneous solutions because this is an odd power. Therefore, –11 is the only solution to the equation.

EXAMPLE 17 Solve the equation 4x+1+ x+2 = 10x+5.

Solution ( 4x+1+ x+2)2 =( 10x+5)2

4x+1+2 (4x+1)(x+2) + x+2 =10x+5 2 (4x+1)(x+2) = 5x+2

(2 (4x+1)(x+2))2 (5x+2)2

4(4x+1)(x+2) 25x2 +20x+4 16x2 +36x+8 = 25x2 +20x+4

9x2 16x 4 = 0 (9x+2)(x 2) = 0

x = 29 or x = 2

Now, check these results in the orijinal equation.

x =	2	;		8	+1+		2	+2 =		20	+5 =	1	+	16	=	5	;	5	=	5
	9			9			9			9		9		9		3		3		3
x = 2 ;			8 +1+ 2 +2 = 20 +5 ; 5 =5

Both statements are true, so both x = 29 and x =2 are solutions.

Equations and Inequalities

267

Check Yourself 11

Solve the equations.

1.		5x+3 = 4					2. 2x 4 = x 2	3. x+2 4x+8 = –3
4.		3x2 2x+15 + 3x2 2x+8 =7
Answers
1.	13		2. 2, 4 3.	7	4.	1	, 1
		5				3

D. EQUATIONS INVOLVING AN ABSOLUTE VALUE

Definition

EXAMPLE 18

Solution

The absolute value of a number is never negative.

|a| 0

On the real number line, the absolute value of x is the distance from the origin to the point x. For example, there are two points whose distance from the origin is three units, –3 and 3. So the equation |x| = 3 has two solutions, 3 and –3. Let us first remember the mathematical definition of absolute value.

absolute value of a function

For all real numbers x,

		f x ,	for	f x 0
f(x)
	=
	– f x ,		for	f x < 0.

We can use this information to begin solving equations involving one or more absolute values.

Solve the equation |x – 2| = 5.
Case 1	Case 2
__________________	_________________
x – 2 0 ; x 2	x – 2 < 0 ; x < 2
x – 2 = 5	–(x – 2) = 5
x = 7	x = –3
So the solutions are –3 and 7.

EXAMPLE 19

Solution 1

Solve |2x – 3| = x + 1.

Case 1

__________________

2x 3 0	; x	3
2x 3 = x+1		2
x = 4
since 4 > 3	, x = 4 is a solution.
2

So the solutions are 4 and 23 .

Case 2

____________________

2x 3 < 0 ;			x <	3
(2x 3) = x+1				2
x =	2	since	2 < 3		,
	3		3	2
x =	2	is a solution.
	3

268	Algebra 11