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

[led]

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

 

[Evgeny.Makarov]

where did i go wrong?

In the very last line. Well, second last line.

 

[MarkFL]

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 $x$:

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

$$x-2=0\implies x=2$$