How do you write 2x^2 - 10 + 3x = (x - 2)^2 + 1 in standard form?

May 14, 2016

${x}^{2} + 7 x - 15 = 0$

Explanation:

Recall that the general equation of a quadratic in standard form is:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} a {x}^{2} + b x + c = 0 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In order to rewrite the given equation in standard form, you need to expand and collect like terms.

Given,

$2 {x}^{2} - 10 + 3 x = {\left(x - 2\right)}^{2} + 1$

Expand and simplify the right side.

$2 {x}^{2} - 10 + 3 x = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\left(x - 2\right) \left(x - 2\right)} + 1$

$2 {x}^{2} - 10 + 3 x = \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{x}^{2} - 4 x + 4} + 1$

$2 {x}^{2} - 10 + 3 x = {x}^{2} - 4 x + 5$

Move all terms to the left side of the equation.

$2 {x}^{2} \textcolor{w h i t e}{i} \textcolor{p u r p \le}{- {x}^{2}} + 3 x \textcolor{w h i t e}{i} \textcolor{p u r p \le}{+ 4 x} - 10 \textcolor{w h i t e}{i} \textcolor{p u r p \le}{- 5} = 0$

Simplify.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{x}^{2} + 7 x - 15 = 0} \textcolor{w h i t e}{\frac{a}{a}} |}}}$