Question

solve 3x^2-32x+12=0 by completing square method

Answer 1

Given: $3 x^{2}-32 x+12=0$

We have to solve this by completing the square method

$3 x^{2}-32 x+12=0\\\\a \neq 1 \text { so divide through by } 3\\\\\frac{3}{3} x^{2}-\frac{32}{3} x+\frac{12}{3}=\frac{0}{3}\\\\\text { which gives us }\\\\x^{2}-\frac{32}{3} x+4=0\\\\x^{2}-\frac{32}{3} x=-4\\\text { Take half of the } x \text { term and square it }\\\left[-\frac{32}{3} \cdot \frac{1}{2}\right]^{2}=\frac{256}{9}$

$\left[-\frac{32}{3} \cdot \frac{1}{2}\right]^{2}=\frac{256}{9}\\\\\text { then add the result to both sides }\\\\x^{2}-\frac{32}{3} x+\frac{256}{9}=-4+\frac{256}{9}\\\\\left(x-\frac{16}{3}\right)^{2}=-4+\frac{256}{9}\\\\\left(x-\frac{16}{3}\right)^{2}=\frac{220}{9}\\\\x-\frac{16}{3}=\pm \sqrt{\frac{220}{9}}$

Therefore,

$\begin{array}{l}x=\frac{16}{3}+\frac{2 \sqrt{55}}{3} \\x=\frac{16}{3}-\frac{2 \sqrt{55}}{3}\end{array}$

Which becomes,

$\begin{array}{c}x=10.2775 \\x=0.389201\end{array}$