# The number of real solutions with the help of discriminant. ### 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 80E
To determine

## The number of real solutions with the help of discriminant.

Expert Solution

The equation has no real roots.

### Explanation of Solution

Given information:

3x2=6x9

Formula used: -

When  b24ac=0  there is one real root.

When  b24ac>0  there are two real roots.

When  b24ac<0  there are two complex roots.

3x2=6x93x26x+9=0

Now, finding discriminant to find number of real solutions:

a=3,b=6,c=9

b24ac=(6)24(3)(9)b24ac=36108b24ac=72

Square root of a negative number is always an imaginary number:

b24ac<0

Hence, no real root exists for equation 3x26x+9=0 .

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