Question

Simplify the expression and solve if needed

Answer 1

\[\frac{ x+3 }{ x^2-5 }=\frac{ 4 }{ x-2 }\]
cross multiply.
\[(x+3)(x-2)=4(x^2-5)\]
\[x^3+x-6=4x^2-20\]
\[0=3x^2-x-14\]
\[0=3x^2+6x-7x-14\]
\[0=3x(x+2)-7(x+2)\]
\[0=(x+2)(3x-7)\]
now you can solve for roots. make each bracket equal to zero.
first bracket:
x+2=0
x=-2
second bracket:
3x-7=0
3x=7
x=7/3
therefore the solutions/roots/x intercepts are x=-2 and x=7/3.

Answer 2

im not sure if thats how you do this

Answer 3

The unit that i am working on is called Multiplying, Dividing, Adding, and subtracting ration expressions

Answer 4

rational*

Answer 5

yeah this is how you do it

Answer 6

hmm what do you mean by brackets though?

Answer 7

which grade are you in? maybe you just haven't gotten this far in your learning yet

Answer 8

11th

Answer 9

Algebra II

Answer 10

oh ok idk what exactly you do in that because im in ib, but do you mean by the ending brackets? when you've factored it, you have 2 brackets. when you divide each of those brackets then in disappears (because on the other side is a 0) and you get 2 solutions for x.

Answer 11

oh ok :P

Answer 12

but in the steps how did you get from
0=3x^2−x−14
to
0=3x^2+6x−7x−14 ??

Answer 13

oh ok. so this is factoring. you're trying to split up the middle term, bx (-1x or -x) into 2 numbers. so which two numbers, when multipled, equal to ab (-42) and those same two numbers, when added, add up to b (or -1)? those two numbers are -6 and +7. so we've split up the middle term.

Answer 14

But i thought you had to factor what went in to -14 and -x

Answer 15

Can someone confirm if this is correct or not cause im not sure if this is how we are supposed to do this

Answer 16

What u are telling is only applicable in the case where the equation is in the form of
ax^2 + bx +c =0
and a is 1
If it is greater than 1 we multiply it with "c" to get a new "c"
and then factor the equation as usual.

Answer 17

He @dg98 is absolutely correct!

Answer 18

no, you're supposed to split up the middle term, as i said, so that you find 2 numbers that equal to ac when multiplied and b when added (same 2 numbers). @╰☆╮Openstudier╰☆╮ , it also works when a=1, because you have to find 2 numbers that equal to ac, and one of them is 1, so ignoring "a" is just a shortcut, my way's just easier to remember for me.

Answer 19

Yeah i know but any number multiplied by equals that that number anyways its the same logic but it is easier to explain the previous one