# Roots of the Equation

• Nov 17th 2008, 10:18 PM
magentarita
Roots of the Equation
The roots of the equation
ax^2 + 4x = –2 are real, rational, and equal when a has a value of

(1) 1 (3) 3
(2) 2 (4) 4

• Nov 18th 2008, 01:54 PM
masters
Quote:

Originally Posted by magentarita
The roots of the equation
ax^2 + 4x = –2 are real, rational, and equal when a has a value of

(1) 1 (3) 3
(2) 2 (4) 4

If a=1, then \$\displaystyle x^2+4x+2=0\$ has irrational roots. (Use quadratic formula)

If a=2, then \$\displaystyle 2x^2+4x+2=0\$. Dividing all terms by 2, we have:

\$\displaystyle x^2+2x+1=0\$. Now factor to get:

\$\displaystyle (x+1)(x+1)=0\$. We have 2 equal roots (double root) at x=-1

If you test a=3 and a-4, you'll see that they have imaginary roots.
• Nov 18th 2008, 09:30 PM
magentarita
ok...........
Quote:

Originally Posted by masters
If a=1, then \$\displaystyle x^2+4x+2=0\$ has irrational roots. (Use quadratic formula)

If a=2, then \$\displaystyle 2x^2+4x+2=0\$. Dividing all terms by 2, we have:

\$\displaystyle x^2+2x+1=0\$. Now factor to get:

\$\displaystyle (x+1)(x+1)=0\$. We have 2 equal roots (double root) at x=-1

If you test a=3 and a-4, you'll see that they have imaginary roots.

A wonderful repy as always.