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 5E
To determine

To fill: The blanks in the statement “The equation (x+1)25(x+1)+6=0 is of ____ type. To solve the equation, we set W=_____. The resulting quadratic equation is _____”.

Expert Solution

Answer to Problem 5E

The complete statement is “The equation (x+1)25(x+1)+6=0 is of quadratic_ type. To solve the equation, we set W=x+1_. The resulting quadratic equation is W25W+6=0_.”

Explanation of Solution

The given equation is (x+1)25(x+1)+6=0.

If the expression x+1 in (x+1)25(x+1)+6=0 is considered as a variable, then the resulting equation will be of the form ax2+bx+c=0 where a=1,b=5and c=6.

Thus, the given equation is a quadratic type equation.

To solve the equation, substitute W for x+1. Then, the equation becomes as follows.

(x+1)25(x+1)+6=0W25W+6=0

Thus, the complete statement is “The equation (x+1)25(x+1)+6=0 is of quadratic_ type. To solve the equation, we set W=x+1_. The resulting quadratic equation is W25W+6=0_.”

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