MHB Therefore, the roots of the given equation are $x=-4$ and $x=2$.

AI Thread Summary
The equation (x+3)^2 = 4x + 17 is solved by first expanding and rearranging it to x^2 + 2x - 8 = 0. The correct factorization of this quadratic yields (x+4)(x-2) = 0, leading to the roots x = -4 and x = 2. A common mistake was misidentifying the signs of the roots during the factoring process. The discussion emphasizes the importance of careful sign management in solving quadratic equations. Ultimately, the roots of the equation are confirmed as x = -4 and x = 2.
led
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Solve (x+3)^2 = 4x+17 where did i go wrong?
(x+3)(x+3 )= 4x+17

x^2 + 3x + 3x + 9 = 4x+17

x^2 + 6x + 9 = 4x + 17

x^2 + 6x + 9 - 4x - 17 = 0

x^2 + 2x - 8 = 0

(x-2)(x+4) <-- USING THE CROSS METHOD View attachment 3985

x= -2, 4 is my cross method working incorrect or something? do explain my mistake
 

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led said:
where did i go wrong?
In the very last line. Well, second last line.
 
You have factored correctly, but your roots have the wrong signs.

Another method to solve this equation would be to write it as:

$$(x+3)^2-4x-17=0$$

$$(x+3)^2-4(x+3)-5=0$$

Now factor the quadratic in $x+3$:

$$\left((x+3)+1\right)\left((x+3)-5\right)=0$$

Combine like terms:

$$(x+4)(x-2)=0$$

Now, to find the actual roots, use the zero-factor property, and equate each factor to zero in turn and solve for $x$:

$$x+4=0\implies x=-4$$

$$x-2=0\implies x=2$$
 
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