# How do you solve the quadratic 3x^2-2x=2x+7 using any method?

Nov 4, 2016

$x = + \frac{7}{3} \mathmr{and} x = - 1$

#### Explanation:

Given:$\text{ } 3 {x}^{2} - 2 x = 2 x + 7$

Subtract $2 x$ and $7$ from both sides

$3 {x}^{2} - 4 x - 7 = 0$

7 is a prime number so does not share any common factors other than 1 with the other coefficients.

Compare to the standardised equations:

$y = a {x}^{2} + b x + c \text{ ; } x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where$\text{ "a=3"; "b=-4"; } c = - 7$ giving:

$x = \frac{+ 4 \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(3\right) \left(- 7\right)}}{2 \left(3\right)}$

$x = \frac{2}{3} \pm \frac{\sqrt{16 + 84}}{6}$

$x = \frac{2}{3} \pm \frac{10}{6}$

$\implies x = + \frac{7}{3} \mathmr{and} x = - 1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Nov 4, 2016

Same solution but using completing the square. The process introduces an error that has to be corrected by the inclusion of $k$
Takes longer to explain than do the maths.

#### Explanation:

Given:$\text{ } 3 {x}^{2} - 4 x - 7 = 0$............................Equation(1)

$\textcolor{b l u e}{\text{Completing the square}}$

Write as: $3 \left({x}^{2} - \frac{4}{3} x\right) - 7 = 0$

Let the constant of correction be $k$

At this point $k = 0$

Write as: $3 \left({x}^{2} - \frac{4}{3} x\right) - 7 + k = 0$

Take the square from ${x}^{2}$ outside the brackets

$3 {\left(x - \frac{4}{3} x\right)}^{2} - 7 + k = 0$

Halve the coefficient of $x$ so $- \frac{4}{3} x$ becomes $- \frac{4}{6} x$

$\text{The "4/6" is not simplified on purpose}$

$3 {\left(x - \frac{4}{6} x\right)}^{2} - 7 + k = 0$

Now get rid of the $x$ from $- \frac{4}{6} x$

$\textcolor{red}{3} {\left(x \textcolor{g r e e n}{- \frac{4}{6}}\right)}^{2} - 7 + k = 0$............................Equation(2)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We now have to find the value of $k$ so that the overall value of equation(2) is the same as that of equation(1)

The error comes from the term $\textcolor{red}{3} {\left(\textcolor{g r e e n}{- \frac{4}{6}}\right)}^{2}$ which is additional to that in the original equation(1).

Equation(1)$\to 3 {x}^{2} - 4 x - 7 = 0$
Equation(2)$\to 3 {x}^{2} - 4 x \textcolor{m a \ge n t a}{+ \frac{{\cancel{16}}^{4}}{{\cancel{12}}^{3}}} - 7 + k = 0$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{w h i t e}{.}$

So from equation(2) we have

$\textcolor{red}{3} {\left(\textcolor{g r e e n}{- \frac{4}{6}}\right)}^{2} + k = 0 \text{ "=>" "k=-4/3larr" gets rid of the error}$

So equation(2) becomes:

$3 {\left(x - \frac{2}{3}\right)}^{2} - 7 - \frac{4}{3} = 0$

$3 {\left(x - \frac{2}{3}\right)}^{2} - \frac{25}{3} = 0 \leftarrow \text{ completed square}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solving for } x}$

$\implies {\left(x - \frac{2}{3}\right)}^{2} = \frac{25}{9}$

$x = \frac{2}{3} \pm \sqrt{\frac{25}{9}}$

$x = \frac{2}{3} \pm \frac{5}{3}$

$x = \frac{7}{3} \mathmr{and} x = - 1$