Question

solve the quadratic equation by completing square
$\sf 6x^2+11x+3=0$​

Answer 1

Answer:

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 6x^2+11x+3 =0$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 6x^2+11x=-3$

• multiply both sides by 2a

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 24 (6x^2+11x)=24 (-3)$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 144x^2+264x=-72$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow (12x)^2+2.12x.11=-72$

• Adding b^2 both sides

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow (12x)^2+2.12x.11+(11)^2=(11)^2-72$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow (12x+11)^2=121-72$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow (12x+11) ^2=49$

• taking square roots

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \sqrt {(12x+11)^2}=\sqrt {49}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 12x+11=\underline{+}49$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow 12x=\underline{+}49-11$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {\underline{+}49-11}{12}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {49-11}{12}\quad or\quad x=\dfrac {-49-11}{12}$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=\dfrac {38}{12}\quad or\quad x=\dfrac {-60}{12}$

$\\\qquad\quad\displaystyle\sf\underline{\boxed{\blue {\bf {:}\longrightarrow x=\dfrac {19}{12}\quad or -5}}}$