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# help

0
21
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Solve the quadratic x^2 - 10x + 9 = 0.

May 16, 2020

#1
+1498
+2

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
0

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