Question

I need help with question number 2 and 3. But 2 if I can only get help with one question.

Answer 1

Number 2:

To solve the quadratic equation by using the factorization method;

$x^2-6x+8=0$

We note here that the factors of 8 that can also add up to -6 are -4 and -2.

Therefore we'll have;

$\begin{gathered} x^2-4x-2x+8=0 \\ (x^2-4x)-(2x+8)=0_{} \\ x(x-4)-2(x-4)=0 \\ \text{Collect like terms and we'll have;} \\ (x-4)(x-2)=0 \end{gathered}$

Therefore, we now have;

$\begin{gathered} x-4=0,x-2=0 \\ x=4,x=2 \end{gathered}$

To solve the quadratic equation by using the quadratic equation formula, we would have;

$\begin{gathered} x^2-6x+8=0 \\ \text{The formula is;} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \text{Where;} \\ a=1,b=-6,c=8 \\ x=\frac{-\lbrack-6\rbrack\pm\sqrt[]{(-6^2)-4(1)(8)}}{2(1)} \\ x=\frac{6\pm\sqrt[]{36-32}}{2} \\ x=\frac{6\pm\sqrt[]{4}}{2} \\ x=\frac{6\pm2}{2} \\ \text{Hence, we now have;} \\ x=\frac{6+2}{2},x=\frac{6-2}{2} \\ x=\frac{8}{2},x=\frac{4}{2} \\ x=4,x=2 \end{gathered}$

Therefore, using the quadratic equation formula, we also have;

$x=4,x=2$