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countonme123
Solve quadratic equation by completing the square. Check both roots. problem below
\[2x^2-12x +16=0\]
please explain each step I am having trouble with it
2(x^2-6x+8)=0 2[(x-4)(x-2)]=0
please do explain why you do each step
1st step- Factor out 2. Second step- What 2 numbers add to the end value, and multiply to the middle value?
Those 2 values are then put into the brackets as you see them.
And here, the signs of positive and negatives are reversed because when you solve for x individually, x-4=0 x=4. Ok?
ok but there is supposed to be two answers
eh, you might be might be making this more difficult than it actually is
I did what the question asked. Solve quadratic equation by completing the square.
You made a small mistake. The -8 should also be multiplied by -2.
Leaving you with\[2(x^2-6x+9-9+8)=2((x-3)^2-1)=2(x-3)^2-2\]
Set equal to 0, and get \[2(x-3)^2-2=0\implies(x-3)^2=1\]Take the root, and solve. \[x-3=\pm1\implies x=4\quad\text{or}\quad x=2\]
ok george you are right now can you explain each step
The hardest part is the first couple steps. You start with \[2x^2-12x+16=2(x^2-6x+8)\]The next part is where you complete the square. You take half of \(-6\), square it, and then add/subtract it. So you get\[2(x^2-6x+8)=2(x^2-6x+\left(\frac{-6}{2}\right)^2-\left(\frac{-6}{2}\right)^2+8=2(x^2-6x+9-9+8)\]
Now you simplify as I described before. It's important that you simplify \[x^2-6x+9-9+8\]as \((x-3)^2-1\). That's the whole point of completing the square. You add/subtract just the right amount so you can have the square of a binomial. From there, you can solve for that square easily, and then just take a square root.
Did this make sense?
yes thank you but please view my next question