x = -7 and x = 1
Explanation:[tex]\begin{gathered} \text{Given:} \\ (x+3)^2\text{ - 14 = 2} \end{gathered}[/tex]
To solve the question, first we will expand the parenthesis:
[tex]\begin{gathered} (x\text{ + 3)(x + 3) - 14 = 2} \\ x(x\text{ + 3) + 3(x + 3) - 14 = 2} \\ \text{the value out is used to multiply the ones in the parenthesis:} \\ x(x)\text{ + x(3) + 3(x) + 3(3) - 14 = 2} \end{gathered}[/tex][tex]\begin{gathered} x^2\text{ + 3x + 3x + 9 - 14 = 2} \\ \text{simplify:} \\ x^2\text{ + 6x - 5 = 2} \\ \text{subtract 2 from both sides:} \\ x^2\text{ + 6x - 5 - 2 = 0} \\ x^2\text{ + 6x - 7 = 0} \end{gathered}[/tex]
Next we'll find the value(s) of x by factorising the quadratic equation:
from the equation above, a = 1, b = 6, c = -7
To factorise using factorisation method we will find the factors of (a × c) whose sum gives b
a × c = 1 × -7 = -7
That is the factors of -7 whose sum gives 6
factors of -7:
1 and -7, -1 and 7
The only factor when you sum them that gives 6 is -1 and 7
so 6x = -x + 7x
factorising, the quadratic equation becomes:
[tex]\begin{gathered} x^2\text{ - x + 7x }-\text{ 7 = 0} \\ x(x\text{ - 1) + 7(x - 1) = 0} \\ (x\text{ + 7)(x - 1) = 0} \\ x\text{ + 7 = 0 or x - 1 = 0} \\ x\text{ = -7 or x = 1} \end{gathered}[/tex]
Hence, the solution is x = -7 and x = 1
