Chapter 9.5, Problem 57E

a.

To determine

To find the value of c for which the equation has two solutions.

Answer to Problem 57E

The value of c should be greater than 1.

Explanation of Solution

Given:

Equation:

  $x^2-2x+c=0$

Calculation:

On comparing the equation with $a{x}^2+bx+c$

  $a=1,b=-2$

Finding the value of c.

For equation to have two solution discriminant must be greater than 0

Therefore,

  $b^2-4ac$ >0

Substituting the values,

  ${(-2)}^2-4(1)c>0$

Opening brackets,

  $4-4c>0$

So,

  $-4c>-4$

Solving root,

  $c>\frac{-4}{-4}$

So,

  $c>1$

Therefore, the value of c should be greater than one.

b.

To determine

To find the value of c for which the equation has one solutions.

Answer to Problem 57E

The value of c should be equal 1.

Explanation of Solution

Given:

Equation:

  $x^2-2x+c=0$

Calculation:

On comparing the equation with $a{x}^2+bx+c$

  $a=1,b=-2$

Finding the value of c.

For equation to have two solution discriminant must equal to 0

Therefore,

  $b^2-4ac$ =0

Substituting the values,

  ${(-2)}^2-4(1)c=0$

Opening brackets,

  $4-4c=0$

So,

  $-4c=-4$

Solving root,

  $c=\frac{-4}{-4}$

So,

  $c=1$

Therefore, the value of c should be equal to 1.

c.

To determine

To find the value of c for which the equation has no solutions.

Answer to Problem 57E

The value of c should be less than 1.

Explanation of Solution

Given:

Equation:

  $x^2-2x+c=0$

Calculation:

On comparing the equation with $a{x}^2+bx+c$

  $a=1,b=-2$

Finding the value of c.

For equation to have two solution discriminant must be less than 0

Therefore,

  $b^2-4ac$<0

Substituting the values,

  ${(-2)}^2-4(1)c<0$

Opening brackets,

  $4-4c<0$

So,

  $-4c<-4$

Solving root,

  $c<\frac{-4}{-4}$

So,

  $c<1$

Therefore, the value of c should be lesser than one.