What two numbers add to be 6 and multiply to be 15

1 answer:

4 0

$\displaystyle \bf \\
x_1+x_2=6\\\\
x_1\cdot x_2=15\\\\
\Longrightarrow~~x^2-6x+15=0\\\\
x_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{6\pm\sqrt{36-60}}{2}=\\\\
=\frac{6\pm\sqrt{-24}}{2}=\frac{6\pm2\sqrt{-6} }{2}=3\pm i\sqrt{6}\\\\
x_1=\boxed{3-i\sqrt{6}}\\\\
x_2=\boxed{3+i\sqrt{6}}\\\\
\texttt{Verify:}\\\\
x_1+x_2=3-\underline{i\sqrt{6}}+3+\underline{i\sqrt{6}}=3+3=6~~correct\\\\
x_1\cdot x_2=(3-i\sqrt{6})(3+i\sqrt{6})=3^2-(i\sqrt{6})^2=9+6=15~~correct$

You might be interested in

A rectangular box has length x and width 3. The volume of the box is given by y = 3x(8 – x). The greatest x-intercept of the gra

CaHeK987 [17]

Consider rectangular box with

length x units (x≥0);
width 3 units;
height (8-x) units (8-x≥0, then x≤8).

The volume of the rectangular box can be calculated as

$V_{box}=\text{length}\cdot \text{width}\cdot \text{height}.$

In your case,

$V_{box}=3\cdot x\cdot (8-x).$

Note that maximal possible value of the height can be 8 units (when x=0 - minimal possible length) and the minimal possible height can be 0 units (when x=8 - maximal possible length).

From the attached graph you can see that the greatest x-intercept is x=8, then the height will be minimal and lenght will be maximal.

Then the volume will be V=0 (minimal).

Answer: correct choices are B (the maximum possible length), C (the minimum possible height)