Question

Find the roots of the equation below. x2 - 6x + 12 = 0

Answer 1

x = 0
(what i first thought)
lets figure this out.
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0*2 - 6*0 + 12 = 0
0 - 0 + 12
12
this is sadly wrong.
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What we need to do, is somehow, make the first two parts of the equation (x2 - 6x) equal -12 to counter the +12.
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maybe x = -2?
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-2 * 2 - 6 * -2 + 12 = 0
-4 - 6 * -2 + 12 = 0
-4 - -12 + 12 = 0
8 + 12 = 20
this is also wrong.
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As we saw in the third step, we were close to getting -12.
Lets try -4.
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-4 * 2 - 6 * -4 + 12 = 0
-8 - 6 * -4 + 12 = 0
-8 - -24 + 12 = 0
16 + 12 = 28
Drat, not close enough.
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Lets try now -3
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-3 * 2 - 6 * -3 + 12 = 0
-6 - 6 * -3 + 12 = 0
-6 - -12 + 12 = 0
-24 + 12 = -12
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now i a frustrated.
by now i realized i was doing it all wrong and now have the answer
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x = 3.
Refine x * 2 - 6 * x + 12 ---> -4x + 12
( 2 - 6 = -4) (x's combine)
-4x + 12
subtract 12 from both sides
-4x - 12 + 12 = 0 - 12
Refine
-4x = -12
divide both sides by 4
-4x / -4 = -12 / -4
x = 3

Answer 2

Answer: The required roots of the given quadratic equation are

`x=3+\sqrt3i,~~~3-\sqrt3i.`

Step-by-step explanation: We are given to find the roots of the following quadratic equation :

`x^2-6x+12=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)`

We will be using the method of completing the square to solve the given quadratic equation (i).

From equation (i), we have

`x^2-6x+12=0\\\\\Rightarrow (x^2-2* x*3+3^2)+12-3^2=0\\\\\Rightarrow (x-3)^2+12-9=0\\\\\Rightarrow (x-3)^2+3=0\\\\\Rightarrow (x-3)^2=-3\\\\\Rightarrow x-3=\pm√(-3)\\\\\Rightarrow x-3=\pm\sqrt3i\\\\\Rightarrow x=3\pm\sqrt3i.`

Thus, the required roots of the given quadratic equation are

`x=3+\sqrt3i,~~~3-\sqrt3i.`