Question

Algebra equation challenge!

Answer 1

\[\frac{ 2x }{ 2x^{2}-5x+3} + \frac{ 13x }{ 2x^{2}+x+3} = 6\]
If you get stuck I'll show you the next step in solving, but medal to the first who finds the answer!

Answer 2

@amistre64 I'm sure you know how to solve it, but please let others try to guess first lol

Answer 3

x=2??

Answer 4

@AnyTipzWouldHelp Do you know how to solve this equation?

Answer 5

Does x=2??

Answer 6

Should I give you the answer? lol

Answer 7

Yes, I know how to solve it, but I want to see if anyone else can, the first who can gets a medal.

Answer 8

Here's a hint: x equals a negative number.

Answer 9

\[(x=- \sqrt{23})\] ????

Answer 10

Dear AnyTip: It is crucial that you find the Lowest Common Denominator (LCD) of\[\frac{ 2x }{ 2x^{2}-5x+3} + \frac{ 13x }{ 2x^{2}+x+3} = 6\]

Answer 11

before attempting to combine the 2 fractions on the left. Hint: Factor each denominator. Then think about how you might obtain the LCD from the various factors of these 2 denominators.

Answer 12

I know, I'm just leaving it open for the first person to say the right answer, then ill give a medal. But I think they are stumped, so thanks for giving them advice lol

Answer 13

And no Yana, that is not the answer.

Answer 14

The answer, rounded to the nearest hundredth, is x=−3.37

Answer 15

\(\dfrac{ 2x }{ 2x^{2}-5x+3} + \dfrac{ 13x }{ 2x^{2}+x+3} = 6\)

\(\dfrac{ 2x }{ (2x - 3)(x - 1)} + \dfrac{ 13x }{ 2x^2 + x+3} = 6\)

\((2x - 3)(x - 1)(2x^2 + x+3)\dfrac{ 2x }{ (2x - 3)(x - 1)} +(2x - 3)(x - 1)(2x^2 + x+3) \dfrac{ 13x }{ 2x^2 + x+3} =\)
\( =(2x - 3)(x - 1)(2x^2 + x+3) 6\)


\(\cancel{(2x - 3)}\cancel{(x - 1)}(2x^2 + x+3)\dfrac{ 2x }{ \cancel{(2x - 3)}\cancel{(x - 1)}}+\)
\( +(2x - 3)(x - 1)\cancel{(2x^2 + x+3)} \dfrac{ 13x }{ \cancel{2x^2 + x+3}} =\)
\( =(2x - 3)(x - 1)(2x^2 + x+3) 6\)

\( (2x^2 + x+3)2x + (2x^2 - 5x + 3)13x = (2x^2 + x + 3)(2x^{2}-5x+3)6 \)

\(4x^3 + 2x^2 + 6x + 26x^3 - 65x^2 + 39x =\)
\(= (4x^4 - 10x^3 + 6x^2 + 2x^3 - 5x^2 + 3x + 6x^2 - 15x +

9)6\)

\(30x^3 - 63x^2 + 45x = 24x^4 - 48x^3 + 42x^2 - 72x + 54\)

\(24x^4 - 78x^3 + 105x^2 - 117x + 54 = 0\)

\(8x^4 - 26x^3 + 35x^2 - 39x + 18 = 0\)

Now let's graph this 4th degree polynomial to get an idea of where to look for real zeros.

Answer 16

Without graphing, we can see the following. As x becomes less than -10, y will continue to get larger, so there are no zeros there. The same is true for x greater than 10.

We found a zero at x = 2.
We can also tell that there is a zero at 0 < x < 1.

Answer 17

Now let's look at the Descartes rule of signs:

f(x) = 8x^4 - 26x^3 + 35x^2 - 39x + 18
f(-x) = 8x^4 + 26x^3 + 35x^2 + 39x + 18

There are 4 or 2 positive roots.
There are no negative roots.

Since from the table above we know there is a root at x = 2 and a root at 0 < x < 1, we know there are 2 positive roots. The other two roots must be complex.

Answer 18

Now we use the Rational Zeros Theorem to help us.

Since we know we have a root at x = 2, let's use synthetic division to bring the polynomial down by one degree.



Now we need to find a zero between 0 and 1 of

8x^3 - 10x^2 + 15x - 9 = 0