solve by completing the square.

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solve by completing the square.

Mathematics
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x\[x ^{2}+2x=24\]
move all one side you get\[x^{2} +2x -24 =0\]
now you factour(x+6)(x-4)

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x=-6 x=4
\[2x ^{2}-4x-3=0\]
this is quadratic equation form: [-b±√(b^2-4(a)(c))]/(2(a)) a=2 b=-4 c=-3 [-(-4)±√((-4)^2-(4*2*(-3))]/(2*2)) [(4)±√(16+24)]/(4) [(4)±2√(10)]/(4) x =1/2[±√(10)]
The above method is not "completing the square"
annette the second you want do complete square?
Even in first one, it is splitting the middle term method and not completing the square
Thank you Harkirat for remind me let do again , can you do second one it late I have to go 2/2=1 , 1^2 =1 (x+1)^2=24+1 (x+1)^2=25 x +1 = √(25) x = -1± 5 x=-6 or x= 4
For the second we proceed as follows: \[2x ^{2}-4x-3=0\]
First we take -3 to RHS as 3\[2x ^{2}-4x=3\]
Next we divide both sides by 2 (to get rid of the 2 in front of x^2) So we get \[x ^{2}-2x=3/2\]
Now to complete the square we add 1^2 to both sides \[x ^{2}-2x+1^{2}=3/2 + 1\]
Now left side is a complete square as follows \[(x-1)^{2}= 5/2\] This gives \[x-1 = \pm \sqrt{5/2}\] Now taking -1 to RHS we get the final answer \[x = 1 \pm \sqrt{5/2}\]
So the two roots/zeros of the given quadratic equation are \[1+\sqrt{5/2}\] and \[1-\sqrt{5/2}\]

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