Answer:
[tex]\textsf{Rewrite the original equation as $x^2+\dfrac{1}{3}x=\boxed{\dfrac{2}{9}}$}[/tex]
[tex]\textsf{Add appropriate number to make the left side a perfect square trinomial}[/tex]
[tex]x^2+\dfrac{1}{3}x+\boxed{\dfrac{1}{36}}=\dfrac{2}{9}+\boxed{\dfrac{1}{36}}[/tex]
[tex]\textsf{Factor the left side as a perfect square and combine the right hand side into one number}[/tex][tex]\left(x+\boxed{\dfrac{1}{6}}\:\right)^2=\boxed{\dfrac{1}{4}}[/tex]
[tex]\textsf{Final answers $x=\boxed{\dfrac{1}{3}, - \dfrac{2}{3}}$}[/tex]
Step-by-step explanation:
Given equation:
[tex]18x^2+6x-4=0[/tex]
Add 4 to both sides:
[tex]\implies 18x^2+6x-4+4=0+4[/tex]
[tex]\implies 18x^2+6x=4[/tex]
Divide both sides by 18:
[tex]\implies \dfrac{18x^2}{18}+\dfrac{6x}{18}=\dfrac{4}{18}[/tex]
[tex]\implies x^2+\dfrac{1}{3}x=\dfrac{2}{9}[/tex]
Add the square of half the coefficient of x to both sides:
[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{\frac{1}{3}}{2}\right)^2=\dfrac{2}{9}+\left(\dfrac{\frac{1}{3}}{2}\right)^2[/tex]
[tex]\implies x^2+\dfrac{1}{3}x+\left(\dfrac{1}{6}}\right)^2=\dfrac{2}{9}+\left(\dfrac{1}{6}\right)^2[/tex]
[tex]\implies x^2+\dfrac{1}{3}x+\dfrac{1}{36}=\dfrac{2}{9}+\dfrac{1}{36}[/tex]
Factor the perfect square trinomial on the left side and combine the numbers on the right side:
[tex]\implies \left(x+\dfrac{1}{6}\right)^2=\dfrac{1}{4}[/tex]
Square root both sides:
[tex]\implies \sqrt{\left(x+\dfrac{1}{6}\right)^2}=\sqrt{\dfrac{1}{4}}[/tex]
[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{\sqrt{1}}{\sqrt{4}}[/tex]
[tex]\implies x+\dfrac{1}{6}=\pm \dfrac{1}{2}[/tex]
Subtract 1/6 from both sides:
[tex]\implies x+\dfrac{1}{6}-\dfrac{1}{6}=\pm\dfrac{1}{2}-\dfrac{1}{6}[/tex]
[tex]\implies x=-\dfrac{1}{6}\pm\dfrac{1}{2}[/tex]
Therefore:
[tex]\implies x=-\dfrac{1}{6}+\dfrac{1}{2}=\dfrac{1}{3}[/tex]
[tex]\implies x=-\dfrac{1}{6}-\dfrac{1}{2}=-\dfrac{2}{3}[/tex]
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