BuyFind

Precalculus: Mathematics for Calcu...

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

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

Answer to Problem 124E

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

Answer to Problem 124E

Both the methods give same solution.

Explanation of Solution

Given: x-x-2=0 .

Method 1.

Get a common denominator;

Use quadratic formula:

  x=-b±b2-4ac2a .

Method 2.

Use direct quadratic formula:

  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;

Use quadratic formula:

  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.

Use direct quadratic formula:

  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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