# To prove: that theone equation in different ways using different given method gives a same solution.

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

### Precalculus: Mathematics for Calcu...

6th Edition
Stewart + 5 others
Publisher: Cengage Learning
ISBN: 9780840068071

#### Solutions

Chapter 1.5, Problem 124E

(a)

To determine

## To prove:that theone equation in different ways using different given method gives a same solution.

Expert Solution

Both solutions are same and equal.

### Explanation of Solution

Given:

x-x-2=0 .

Method 1:

Substituting method

letting x=u .

Method 2:

Direct solving method.

Concept used:

The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where x=-b±b2-4ac2ax=-b±b2-4ac2a .

Calculation:

(a)

Method 1:

x-x-2=0 .

Consider the u=x .

u2=x .

x-x-2=0 .

u2-u-2=0 .

(u-2)(u+1)=0 .

u=2,-1

x=4,x=1 .

Method 2:

x-x-2=0 .

x-2=x .

Squaring both side:

x2-4x+4=x .

x2-5x+4=0 .

(x-4)(x-1)=0 .

x=4,x=1 .

Hence both solutions are same and equal.

(b)

To determine

### To prove:that the one equation in different ways using different given method gives a same solution.

Expert Solution

Both the methods give same solution.

### Explanation of Solution

Given: x-x-2=0 .

Method 1.

Get a common denominator;

x=-b±b2-4ac2a .

Method 2.

x=-b±b2-4ac2a .

Concept used:

The First it can solve the given equation through middle term factorization by taking common factor and if it doesn’t get factorize then use discriminant method to solve the given quadratic equation where x=-b±b2-4ac2a .

Calculation:

(b) .

Method 1.

Get a common denominator;

x=-b±b2-4ac2a .

12(x-3)2+10(x-3)+1=0 .

It can be written as:

12(x-3)2+10(x-3)(x-3)2+(x-3)2(x-3)2=0 .

12+10(x-3)+(x-3)2=0 .

x2+4x-9=0 .

x=-4±(4)2-4(1)(9)2(1) .

x=-4±2132(1)=-2±13 .

Method 2.

x=-b±b2-4ac2a

Let u=x-3

12(x-3)2+10(x-3)+1=0 .

12u2+10u+1=0 .

12u2+10uu2+u2u2=0 .

12+10u+u2=0 .

u=-10±(10)2-4(1)(12)2(1) .

u=-10±522(1)=-10±2132(1) .

u=-10±2132=5±13 .

x-3=-5±13 .

x=-2±13 .

Hence both the methods give same solution.

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