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3 years ago

Find the sum and product of the roots. 2x^2 + 6x - 8 = 0

1 answer:

3 0

Suppose you can factorize:

$ax^2+bx+c=a(x-r_1)(x-r_2)$

Then by expanding the right side, you have

$ax^2+bx+c=a(x^2-(r_1+r_2)x+r_1r_2)=ax^2-a(r_1+r_2)x+ar_1r_2$

which means in this case, you have to have

$\begin{cases}6=-2(r_1+r_2)\\-8=2r_1r_2\end{cases}\implies\begin{cases}r_1+r_2=-3\\r_1r_2=-4\end{cases}$

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2 years ago

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2 years ago

I NEED HELP ASAP PLEASE

Ahat [919]

Answer: Domain: 0,1,2,3 and 4 and Range: 1,2,4,8 and 4

Step-by-step explanation:

The domain is the set of all inputs that makes the function definable. As we see the function looks like an exponential function which means the domain is all real numbers.

So the inputs are 0,1,2,3 and 4

the outputs are 1,2,4,8 and 4

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3 years ago

Let us have four distinct collinear points $a,$ $b,$ $c,$ and $d$ on the cartesian plane. the point $c$ is such that $\dfrac{ab}

Deffense [45]

Start with a line segment connecting two points, A and B. $\dfrac{DA}{BA}=3$ means DA is 3 times longer than BA. Clearly, D cannot fall between A and B because that would mean DA is shorter than BA. So there are two possible locations where D can be placed on the line relative to A and B.

But with $\dfrac{DB}{BA}=2$ , or the fact that DB is 2 times longer than BA, we can rule out one of these positions; referring to the attachment, if we place D to the left of A, then DB would be 4 times longer than BA.

Finally, $\dfrac{AB}{CB}=\dfrac12$ , so that CB is 2 times longer than AB. Again we have two possible locations for point C (it cannot fall between A and B), but one of them forces C to occupy the same point as D. However, A, B, C, D are distinct, so C must fall to the left of A.

Now let $d$ be the length of AB. Then the length of CD in terms of $d$ is $4d$ . We have the coordinates of C and D, and the distance between them is $\sqrt{(4-0)^2+(0-4)^2}=4\sqrt2$ . So

$4d=4\sqrt2\implies d=\sqrt2$

The slope of the line through C and D is

$\dfrac{0-4}{4-0}=-1$

and so the equation of the line through these points is

$y-4=-(x-0)\implies x+y=4$

So the coordinates of A are $(x,y)=(x,4-x)$ . The distance between C and A is $d=\sqrt2$ , so we have

$\sqrt{(x-0)^2+(4-x-4)^2}=\sqrt{2x^2}=|x|\sqrt2=\sqrt2\implies|x|=1$

Since A falls to the right of C (in the $x,y$ plane, not just in the sketch), we know to take the positive value $x=1$ . Then the $y$ coordinate is $y=4-1=3$ .

All this to say that A is the point (1, 3), so

$2x+y=2+3=5$

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3 years ago