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# Different methods for solving quadratic equations

6.5  Equations of the Form ax2+bx+c=0

By a quadratic equation in the single variable x, we mean any equation that can be transformed through elementary transformations to an equation of the form

ax2+bx+c=0,  a0

When the equation is written in the above form we will say that it is in standard form.

The ﬁrst type of quadratic equation that we will consider is an equation of the form

ax2+c=0

We first solve for x2 to obtain

x2= ca

Next we solve for x by taking the square root to obtain

x=± ca

The solution set is therefore {( ca), ( ca)}.

Example 1.  Solve 9 x2-25=0

9 x2-25=0

x2=259

x=± 259

x=± (53)

The solution set is {53, 53}.

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6.6  Solution of ax2+bx+c=0 by Factoring

We recall from Chapter 1 the important fact that if A and B are real numbers and

AB=0

then

A=0  or  B=0

This property is the key to solving quadratic equations by the method of factoring. It is important to note that the above reasoning applies only when the right-hand side is zero; no other real number will do.

Example 1.  Solve the equation x2-4 x-5=0.

x2-4 x-5=0

(x-5) (x+1)=0

Therefore,

x-5=0  or x+1=0

so the solution set is {5, 1}

Example 2  Solve the equation 2 x2-5 x=3.

We add 3 to both sides to obtain 0 on the right-hand side.

2 x2-5 x=3

2 x2-5 x-3=0

(2 x+1) (x-3)=0

Therefore,

2 x+1=0  or  x-3=0

so the solution set is

{ 12,3}

More generally if

AB C=0

then

A=0,B=0,  or C=0

Consequently this method of solution will apply to any degree equation as long as we are able to factor the equation into linear terms. For example, if we have

x (x-3) (x+1)=0

then

x=0x-3=0, or x+1=0

and hence the solution set is

{0,3, 1}

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6.7  Completing the Square

As we have seen, quadratic equations of the form

x2=a

where amay be solved by the method of extraction of roots. In fact, since every real number a has two square roots (either real or imaginary) we have two solutions. These are given by is any real number,

x=a  or  x= a

For example , from the equation

x2=25

we have

x=25=5  or  x= 25= 5

From the equation

x2= 9

we have

x= 9=3 1=3 i  or  x= 9= 3 i

Equations of the form

(x-a)2=b

can be solved by the same method. For example

(x-3)2=25

implies that

x-3=5  or  x-3= 5

from which we obtain

x=8  or  x= 2

In order to write x2±2 ax+b=0 in the form

(x±a)2=k

we note that

(x±a)2=x2±2 ax+a2

Hence, in order to obtain a perfect square on the left-hand side when the ﬁrst two terms x2±2 ax are given, we must add a2, which is the square of half the coefficient of the x term, to both sides of the equation. This method is called the method of completing the square. Note that to complete the square the coefficient of x2 must be 1.

Example 1.  Solve the equation x2-10 x-24=0by completing the square. We ﬁrst add 24 to both sides to obtain

x2-10 x+25=24+25

namely,

(x-5)2=49

We now extract the square roots to obtain

x-5=47=7  or  x-5= 49= 7

thus,

x=12  or  x= 2

Therefore, the solution set is {12, 2}.

Example 2.   Solve 4 x2+4 x-3=0.

we add 3 to both sides and then divide by 4 to obtain

x2+x=34

Adding (12)2=14to both sides we obtain

x2+x+14=34+14

or

(x+12)2=1

Extracting roots we obtain

x+12=1  or  x+12= 1.

Thus the solution set is {12,32}.

The method of completing the square has other applications.

This is how our quadratic equation step by step solver solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.

Example 3.  Write x2+y2-4 x+6 y-3=0 in the form (x±h)2+(y±k)2=r2

We group the x terms and the y terms obtaining quadratic expressions in x and y. We then complete the square in each expression.

x2+y2-4 x+6 y-3=0

x2-4 x+y2+6 y=3

x2-4 x+4+y2+6 y+9=3+4+9

(x-2)2+(y+3)2=42

Example 4.  Write y=x2+2 x+3 in the form y±k=(x±h)2.

y=x2+2 x+3

y-3=x2+2 x

y-3+1=x2+2 x+1

y-2=(x+1)2

We can apply the methods of completing the square to the general quadratic equation

ax2+bx+x=0,  a0

ax2+bx= c

Divide sides by a.

x2+bax= ca

Add (b2 a)2=b24 a2 to both sides.

x2+bax+b24 a2= ca+b24 a2

(x+b2 a)2=b2-4 ac4 a2

Take the square root.

x+b2 a

=± b2-4 ac4 a2

=± b2-4 ac2 a

Thus

x= b±sqrt ((b2-4 ac))2 a

This formula, called the quadratic formula, gives both solutions of the general quadratic equation expressed in terms of the coefficients a,b, and c. The symbol ± is used to condense the two equations.

x= b+sqrt ((b2-4 ac))2 ax= b-sqrt ((b2-4 ac))2 a

The solution set of the general quadratic equation is

{x= b+sqrt ((b2-4 ac))2 a,x= b-sqrt ((b2-4 ac))2 a}

In considering the quadratic formula we can see that the nature of the solutions-complex, irrational, or rational-depends entirely on the term under the radical, namely,

b2-4 ac

This term is called the discriminant of the quadratic equation.

1. The solutions are real if b2-4 ac0 and complex if b2-4 ac<0.

2. If a,b, and c are rational and b2-4 ac0, then the solutions are rational if and only if b2-4 ac is a perfect square.

Example 1.  Solve the equation 3 x2+x-2=0 by the quadratic formula.

Substitute a=3b=1,c= 2in the quadratic formula.

x= b±sqrt ((b2-4 ac))2 a

= 1±1-4 (3) ( 2)2·3

= 1±256

= 1±56

=46, 66

Hence the solution set is {23, 1}

Example 2.  Solve the equation x2+4 x+16=0.

We have a=1b=4, and c=16.

x= 4±16-4 (1) (16)2·1

= 4± 482

= 4±4 3 i2

= 2±2 3 i

Thus, the solution set is { 2+2 3 i, 2-2 sqrt (3) i}.

This is how our quadratic equation step by step solver solves the problem above. You can see similar problems solved by clicking on 'Solve similar' button.

In Chapter 3 we were able to factor some polynomials of second degree with integral coefficients. We are now in a position to consider the problem of factoring any second degree polynomial where the coefficients may now be real numbers. We must allow the use of complex numbers in the factors.

Consider the general second degree polynomial and use the method of completing the square.

ax2+bx+c

=a (x2+ba x+ca)

=a (x2+ba x+b24 a2+ca-b24 a2)

=a [(x+b2 a)2-(b2-4 ac4 a2)]

=a [(x+b2 a)2-(b2-4 ac2 a)2]

=a [(x+b2 a)-(b2-4 ac2 a)][(x+b2 a)+b2-4 ac2 a]

=a [x-( b+b2-4 ac2 a)][x-( b-b2-4 ac2 a)]

If we donate  b+b2-4 ac2 a by r1 and  b-b2-4 ac2 a by r2, then we have

ax2+bx+c=a (x-r1) (x-r2)

We observe that the numbers r1, and r2 are just the solutions of ax2+bx+c=0. These solutions, r1 and r2, are called the roots of the polynomial ax2+bx+c. Thus, to factor ax2+bx+c, we obtain its roots by solving the polynomial equation ax2+bx+c=0.

Example 3.  Factor 3 x2+2 x+1.

We ﬁnd the two roots of the polynomial, that is, the solutions of the polynomial equation

3 x2+2 x+1=0

x= 2±4-4 (3) (1)6

= 2± 86

= 1±2 i3

Since the two solutions are  1+2 i3 and  1-2 i3, we have

3 x2+2 x+1=3 (x- 1+2 i3) (x- 1-2 i3)

If an equation in more than one variable is a quadratic equation in one of its variables, then the equation can be solved for that variable by using the quadratic formula.

Example 4.  Solve x2+5+2 y= 7 xy for x.

x2+5+2 y=7 xy

x2-7 xy+(2 y+5)=0

As a quadratic in x we have a=1,b=7 y, and c=2 y+5. Therefore,

x= 7 y±(7 y)2-4 (1) (2 y+5)2

= 7 y±49 y2-8 y-202

By various techniques certain types of equations lead to quadratic equations. Often these techniques produce equations that are not equivalent to the original ones. The steps in the solution of a given problem should be carefully reviewed to see if the replacement set changes at a given step. If so, we must make sure that the solutions of the ﬁnal equation are solutions of the original equation.

Example 1.  Solve the equation x+5x+3-x+6x=23.

Note that x0,3. Multiply by the L.C.D. 3 x (x-3).

3 x (x+5)-3 (x-3) (x+6)=2 x (x-3)

Solve the equation

3 x (x+5)-3 (x-3) (x+6)=2 x (x-3)

3 x2+15 x-3 x2-9 x+54=2 x2-6 x

2 x2-12 x-54=0

x2-6 x-27=0

(x-9) (x+3)=0

The solution set of the last equation is {9, 3}. This is also the solution set of the original equation, since the only numbers not in the replacement set of the original equation are 0 and 3, neither of which is an element of the solution set of the last equation.

Example 2.  Find all real solutions of the equation 2 x+5=x+1.

Square both sides.

2 x+5=(x+1)2

The replacement set of this equation is the set of all real numbers, while the replacement set of the original equation is not the set of all real numbers. Since the operation of squaring enlarges the replacement set, the solution set may also be enlarged. Therefore we must check the solutions that we obtain for the second equation in the original equation.

2 x+5=(x+1)2

2 x+5=x2+2 x+1

x2=4

x=± 2

The possible solution are 2 and  2.

2 (2)+5=9

=3

(2)+1=3

Therefore, 2 is a solution.

2 ( 2)+5=1

=1

( 2)+1= 1

Therefore, 2 is not a solution. The solution set is {2}.

Example 3.  Solve the equation (x2-4)2-5 (x2-4)+6=0.

(x2-4)2-5 (x2-4)+6=0

Let A=x2-4

A2-5 A+6=0

Factor.

(A-3) (A-2)=0

A=3,  A=2

Since A=x2-4

x2-4=3  or  x2-4=2

both of which we now solve.

x2-4=3

x2=7

x=± 7

x2-4=2

x2=6

x=± 6

The solution set is therefore {7, 7,6, 6}.

Example 4.  Find all real solution of Z32-7 Z34-8=0.

Z32-7 Z34-8=0

Let B=Z34,  then B2=Z32.

B2-7 B-8=0

Factor.

(B-8) (B+8)=0

B=8  or  B= 1

Since B=Z34,

Z34=8  or  Z34= 1

Solving, we have

Z34=8

Z=834

=24

=16

Z34= 1

Z=( 1)43

=1

The solution Z=1, however, does not satisfy the equation Z34= 1. In fact we could have observed that this equation has no real solution, since for negative real numbers Z,Z34 is not deﬁned, while for non-negative Z,Z34 is non-negative. It is easy to see that Z=16 does satisfy our original equation, and hence the solution set is {16}.

Example 5.  Solve x2+125=0.

Factor.

x2+125=0

x3+53=0

(x+5) (x2-5 x+25)=0

x+5=0  or  x2-5 x+25=0

Solve.

x+5=0

x= 5

x2-5 x+25=0

x=5±25-4 (1) (25)2 (1)

=5± 752

=5±5 3 i2

Hence the solution set is { 5,52+(5 32) i,52-(5 32) i}.

Example 6.  Solve 2 x+4=x-2.

2 x+4=x-2

Square both sides.

(2 x+4)2=(x-2)2

2 x+4=x-4 x+4

x= 4 x

Square both sides.

x2=( 4 x)2

x2=16 x

x2-16 x=0

x (x-16)=0

x=0  or  x=16

Check.

2 (0)+4=2

0-2= 2

Therefore, 0 is not a solution.

2 (16)+4=36

=6

16-2=4-2

=2

Therefore, 16 is not a solution. Since 0 and 16 are the only possible solutions, the original equation has no solution. In case an equation has no solutions we use the symbol to represent the solution set, that is, the set with no elements. The set is called the empty set.

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6.10  Statement Problems of the Quadratic Type

Our method of approach will be the same as in Section 6.4, only here our equation will be one that yields a quadratic equation in a single variable.

Example 1.  Find two numbers whose sum is 8 and whose product is 12.

Step 2. Let x be one of the numbers.

Step 3. Since their product is 12 the other number is 12x.

Step 4. The sum of the two numbers is 8.

Step 5. x+12x=8

Step 6. Solve.

x2+12=8 x

x2-8 x+12=0

(x-6) (x-2)=0

x=6  or  x=2

In either case the two numbers are 6 and 2.

Example 2. If the perimeter of a rectangle is 44 ft while the area is 120 ft2,what are its dimensions?

Step 2. Let x be the length of one side.

Step 3. Since 12 of the perimeter is 22 ft the other side is 22-x.

Step 4. The area of the rectangle is 120 ft2.

Step 5. x (22-x)=120

Step 6. Solve.

22-x2=120

x2-22 x+120=0

(x-10) (x+12)=0

x=10  or  x=12

In either case the two sides are 10 ft  and 12 ft in length.

Example 3. Within a rectangular garden 18 yards long and 10 yards wide we are to pave a border of uniform width. If we are to leave an area of 84 square yards for ﬂowers, how wide should the walk be?

Step 2. Let x be the width of the walk.

Step 3. The dimensions of the inside area would be 18-2 x and 10-2 x.

Step 4. The inside area is to be 84 sq. yds.

Step 5. (18-2 x) (10-2 x)=84

Step 6. Solve.

180-56 x+4 x2=84

4 x2-56 x+96=0

x2-14 x+24=0

(x-12) (x-2)=0

x=12  or  x=2

Clearly the width of the walk would be 2 yards and not 12 yards.

Example 4. If Joe takes 6 hours longer than his father to assemble a machine and together they can assemble it in 4 hours, how long would it take each of them working alone to assemble the machine?

Step 2. Let x be the number of hours it would take the father to assemble the machine.

Step 3. It would take x+6 hours for Joe to assemble the machine. Choosing one hour as our unit of time, 1x would be the amount the father could do in one hour, while 14 would be the amount that they could do together in one hour.
Step 4. The amount that Joe can do in one hour plus the amount that his father can do in one hour would be equal to 14.

Step 5. 1x+6+1x=14.

Step 6. Solve. Multiply by the L.C.D. 4 x (x+6).

4 x+4 (x+6)=x (x+6)

4 x+4 x+24=x2+6 x

x2-2 x-24=0

(x-6) (x+4)=0

x=6  or  x= 4

Clearly x= 4 is not a possibility and therefore it would take Joe 12 hours and his father 6 hours.

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