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what are the steps for finding the roots of this equation using quadratics: x^2+5x+6=0

Guest Mar 3, 2017

Best Answer 

 #4
avatar+4777 
+11

The fastest way is with factoring.

x2 + 5x + 6 = 0

(x+2)(x+3) = 0

Set each factor equal to zero.

x + 2 = 0       and      x + 3 = 0

x = -2            and      x = -3

 

You can also use the quadratic formula.

\(ax^{2}+bx+c=0 \\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \\ a=1, b=5, c=6 \\ x = {-5 \pm \sqrt{5^2-4(1)(6)} \over 2(1)} \\ x = {-5 \pm \sqrt{1} \over 2} \\ x = \frac{-5+1}{2} \text{ and } x = \frac{-5-1}{2} \\ x=-2 \text{ and } x=-3\)

 

A third method is called completing the square, here's an explanation of that method:

https://www.youtube.com/watch?v=bNQY0z76M5A

hectictar  Mar 3, 2017
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3+0 Answers

 #1
avatar+321 
0

Use the calculator to find out

Davis  Mar 3, 2017
 #4
avatar+4777 
+11
Best Answer

The fastest way is with factoring.

x2 + 5x + 6 = 0

(x+2)(x+3) = 0

Set each factor equal to zero.

x + 2 = 0       and      x + 3 = 0

x = -2            and      x = -3

 

You can also use the quadratic formula.

\(ax^{2}+bx+c=0 \\ x = {-b \pm \sqrt{b^2-4ac} \over 2a} \\ a=1, b=5, c=6 \\ x = {-5 \pm \sqrt{5^2-4(1)(6)} \over 2(1)} \\ x = {-5 \pm \sqrt{1} \over 2} \\ x = \frac{-5+1}{2} \text{ and } x = \frac{-5-1}{2} \\ x=-2 \text{ and } x=-3\)

 

A third method is called completing the square, here's an explanation of that method:

https://www.youtube.com/watch?v=bNQY0z76M5A

hectictar  Mar 3, 2017

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