Question

Solve $3x^2-14x+8=0$

• A. $4$ or $\frac{2}{3}$
• B. $-\frac{2}{3}$ or $-4$
• C. $-4$ or $\frac{2}{3}$
• D. $4$ or $-\frac{2}{3}$

Answer 1

Answer: (C)

$-4$ or $\frac{2}{3}$

Dividing both sides by 3, we get
$x^2-\frac{14}{3}x+\frac{8}{3}=0$
Now, we add to both sides the square of half the coefficient of x to make the left side a perfect square. Adding 49/9 to both sides, we get
$x^2-\frac{14}{x}x+\frac{49}{9}=\frac{49}{9}-\frac{8}{3}$
or ${(x-\frac{7}{3})}^2=\frac{25}{9}$
or $x-\frac{7}{3}=\pm \frac{5}{3}$
i.e., $x=\frac{7}{3}\pm \frac{5}{3}=\frac{12}{3},\frac{2}{3}$
$\therefore x=-4$ or $\frac{2}{3}$