We are given the equation x² + 8x = 33, and we need to solve it from completing the square, for which we will be following the below steps
Step 1 :- First, notice, what's the coefficient of x², if it's 1, then proceed to step 2, if not then divide both sides by the coefficient of x²
[tex]{:\implies \quad \sf x^{2}+8x=33}[/tex]
Step 2 :- Here, we need to develop a whole square in both sides somehow, so we have the coefficient of x, so add the square of half of coefficient of x on both sides :
[tex]{:\implies \quad \sf x^{2}+8x+{\bigg(\dfrac82\bigg)}^{2}=33+{\bigg(\dfrac82\bigg)}^{2}}[/tex]
Simplify both sides now :
[tex]{:\implies \quad \sf x^{2}+8x+(4)^{2}=33+(4)^{2}}[/tex]
Step 3 :- So now, here LHS is in the form of a² + 2ab + b², so we can just replace it by (a + b)², and we will keep simplifying RHS
[tex]{:\implies \quad \sf (x+4)^{2}=33+16}[/tex]
[tex]{:\implies \quad \sf (x+4)^{2}=49}[/tex]
[tex]{:\implies \quad \sf (x+4)^{2}=(\pm 7)^{2}}[/tex]
Step 4 :- As, power on both sides is 2, so we can just equate the bases
[tex]{:\implies \quad \sf x+4=\pm 7}[/tex]
Step 5 :- Equating x + 4 to 7, will yield x = 3, and equating it to -7, will yield x = -11, so we are now left with
[tex]{:\implies \quad \boxed{\bf{x=-11,3}}}[/tex]
