How do you solve #3x^2-4x-5=0# by completing the square?

Question

Nathan Axel · Accepted Answer

First, divide by a = 3
#x^2 - 4x/3 - 5/3 = 0#
# x^2 - 4x/3 + 4/9 - 4/9 - 5/3 = 0# #(x - 2/3)^2 = 4/9 + 5/3 = 19/9# a.# (x - 2/3) = (sqr19)/3#---># x = 2/3 + (sqr19)/3 # b.# (x - 2/3) =(-sqr19)/3 -> x = 2/3 -(sqr19)/3#

Eva Abramson · Answer

undefined To solve $3x^2 - 4x - 5 = 0$ by completing the square, first, move the constant term to the other side to isolate the quadratic terms: $3x^2 - 4x = 5$. Then, divide both sides by the coefficient of the squared term to make the coefficient of $x^2$ equal to 1: $x^2 - \frac{4}{3}x = \frac{5}{3}$. Next, add and subtract $(\frac{1}{2} \cdot -\frac{4}{3})^2 = (\frac{-2}{3})^2 = \frac{4}{9}$ inside the parentheses to complete the square: $x^2 - \frac{4}{3}x + \frac{4}{9} - \frac{4}{9} = \frac{5}{3}$. Rewrite the left side as a perfect square trinomial and simplify the constant terms: $(x - \frac{2}{3})^2 - \frac{4}{9} = \frac{5}{3}$. Add $\frac{4}{9}$ to both sides: $(x - \frac{2}{3})^2 = \frac{5}{3} + \frac{4}{9} = \frac{15}{9} + \frac{4}{9} = \frac{19}{9}$. Take the square root of both sides: $x - \frac{2}{3} = \pm \sqrt{\frac{19}{9}}$. Solve for $x$: $x = \frac{2}{3} \pm \frac{\sqrt{19}}{3}$. Therefore, the solutions are $x = \frac{2 + \sqrt{19}}{3}$ and $x = \frac{2 - \sqrt{19}}{3}$.