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Solve the quadratic x^2 - 10x + 9 = 0.

 May 16, 2020
 #1
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Plugging it in to the quadratic formula

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

We have.

 

\(x = {10 \pm \sqrt{-10^2-4*9} \over 2} = {10 \pm \sqrt{64}\over 2} = {10 \pm 8}\over2\)

 

So 10 +/- 8  OVER 2 is  18/2 or 1. 

 

Plugging into the equation

 

9^2 - 90 +9 = 0 = 81 + 9 - 90 = 0

 

YUP!

 

1^2 - 10 + 9 = 0 = 1 + 9 - 10 = 10-10 = 0.

 

YUP!

 

Great those are our roots.

 

A quadratic can only have two,

 

If you don't understand anything feel free to ask!

 May 16, 2020
 #2
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Solve the quadratic x^2 - 10x + 9 = 0. 

 

Alternatively, this one is simple enought that you could just factor it. 

 

What two numbers multiply together to get +9 and add together to get –10 ? 

 

I wish they were all this easy.  Of course, the numbers are –1 and –9 

 

So,   x2 – 10x + 9 = 0   factored is   (x –1)(x – 9) = 0  

 

Setting each component equal to zero:    (x – 1) = 0 

                                                                 x = 1 

 

                                                                 (x – 9) = 0  

                                                                 x = 9 

.

 May 16, 2020

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