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 84E
To determine

The number of real solutions with the help of discriminant.

Expert Solution

The equation has two distinct real solutions.

Explanation of Solution

Given information:

x2+rxs=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.

x2+rxs=0

Now, finding discriminant to find number of real solutions:

a=1,b=r,c=s

D=b24ac=(r)24(1)(s)=r2+4s,(s>0)=r2+4s

Since D>0, So the equation has two distinct real solutions.

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