Saturday

March 28, 2015

March 28, 2015

Posted by **Haylee** on Sunday, May 13, 2007 at 11:26pm.

#1) x^2+3x+1

#2) x^2+4x+3=0

#3) -2x^2+4x-5

#3 is not an equation. Dod you omit "= 0" at the end?

#2 can be factored into (x+1)(x+3) = 0, so the roots are x=-1 and -3.

#1 Use the quadratic equation for the roots

[-b +/- sqrt (b^2-4ac)]/2a = [-3 +/-sqrt5]/2

"Roots" are solutions to an equation, or those particular values of the variable which make the equation a true statement.

Only your second problem is an equation, I will assume you meant the other two to be equations as well.

There are three common methods to solve quadratic equations:

1. by factoring. Only quadratic equations with rational roots will factor, so they are the exception

2. by completing the square. If the coefficient of the square term is 1, and the middle coefficient is even, this is a very good and fast method.

3. by the quadratic equation formuls. This is a sure method that works for all real and complex roots.

I will solve the #2 equation in those three ways.

x^2 + 4x + 3 = 0

(x+1)(x+3)=0

x+1=0 or x+3=0

so x= -1 or x=-3

x^2 + 4x + 3 = 0

x^2 + 4x +

x^2 + 4x + 4 = -3 + 4

(x + 2)^2 = 1

x+2 = ħ1

x = -1 or x=-3

x^2 + 4x + 3 = 0

x = (-4 ħsqrt(16-4*1*3))/2

x = (-4 ħ 2)/2

x=-1 or x=-3

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