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
![{:\implies \quad \sf x^(2)+8x=33}](https://img.qammunity.org/2023/formulas/mathematics/college/8mwa7cpa3orl0wkdd7a3uqjngvergnk3z4.png)
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 :
![{:\implies \quad \sf x^(2)+8x+{\bigg(\frac82\bigg)}^(2)=33+{\bigg(\frac82\bigg)}^(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/ext52vt53pk4fhy9d41hrks23kf9p9qrx6.png)
Simplify both sides now :
![{:\implies \quad \sf x^(2)+8x+(4)^(2)=33+(4)^(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/jxq68p70fzvstkaskdgty6y0btjemplvqt.png)
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
![{:\implies \quad \sf (x+4)^(2)=33+16}](https://img.qammunity.org/2023/formulas/mathematics/college/8lmtvrpzpy2uyb6x033xjfx2v46np1tcwi.png)
![{:\implies \quad \sf (x+4)^(2)=49}](https://img.qammunity.org/2023/formulas/mathematics/college/azrepc0wyakth6spz8x7qd142ush0c3x0w.png)
![{:\implies \quad \sf (x+4)^(2)=(\pm 7)^(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/2pg9br0sgicblvfz16p6nfgbj9o62mgdpv.png)
Step 4 :- As, power on both sides is 2, so we can just equate the bases
![{:\implies \quad \sf x+4=\pm 7}](https://img.qammunity.org/2023/formulas/mathematics/college/gr1y3w7lm87jyebgfo1gunv8pf28lceiuv.png)
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
![{:\implies \quad \boxed{\bf{x=-11,3}}}](https://img.qammunity.org/2023/formulas/mathematics/college/39nsj90w2eyq5r25pqy4wd48016xgtaemi.png)