# 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 79E
To determine

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

Expert Solution

The equation has two real roots

### Explanation of Solution

Given information:

x26x+1=0

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.

x26x+1=0

Now, finding discriminant to find number of real solutions:

a=1,b=6,c=1

b24ac=(6)24(1)(1)b24ac=364b24ac=32b24ac=5.65

b24ac>0

Hence, there are two real roots of equation x26x+1=0 .

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