How do you solve using the completing the square method #x^2+12x+23=0#?
We have that by applying the square method.
#x^2+12x+23=x^2+26x+6^2+(236^2)= (x+6)^213#
Hence from #x^2+12x+23=0=>(x+6)^213=0=> x+6=+sqrt13=>x_1=6+sqrt13 or x_2=6sqrt13#
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To solve the equation ( x^2 + 12x + 23 = 0 ) using the completing the square method, follow these steps:

Move the constant term to the other side of the equation: ( x^2 + 12x = 23 )

Add and subtract the square of half of the coefficient of ( x ) (12/2 = 6): ( x^2 + 12x + 36  36 = 23 )

Factor the perfect square trinomial and simplify: ( (x + 6)^2  36 = 23 )

Add 36 to both sides of the equation to isolate the perfect square trinomial: ( (x + 6)^2 = 13 )

Take the square root of both sides: ( x + 6 = \pm \sqrt{13} )

Subtract 6 from both sides to solve for ( x ): ( x = 6 \pm \sqrt{13} )
Therefore, the solutions to the equation are ( x = 6 + \sqrt{13} ) and ( x = 6  \sqrt{13} ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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