Question

2x^2 - x + 10 = 0 HELP ME PLEASE!!! I will give a medal!!!!

Answer 1

use the quadratic formula.

Answer 2

yeah, I've got it now thanks!!!

Answer 3

I gave you a medal just for helping me!!! :)

Answer 4

yeah, that was really kind :) Do it and show me what you get, to make sure you're correct.

Answer 5

ok

Answer 6

hold on, are you familiar with \(\color{blue}{ i }\) ?

Answer 7

????

Answer 8

No? Because you will get a negative root if you do the quadratic formula.

Answer 9

here are the solutions!!
-2, 2.5
-1.97, 2.47
-2.5, 2
no solution

Answer 10

\(\color{blue}{ i =\sqrt{-1}}\)

Answer 11

Which course are you taking ?

Answer 12

Algebra 1

Answer 13

If you are taking trig, then say \(\huge\color{blue}{ \frac{2±i\sqrt{79}}{4} }\)

but for algebra 1, it would be just no solution, because negative root is no solution (for alg 1 )

Answer 14

yeah. it would be no solush.

Answer 15

oh ok thank you!!!

Answer 16

You welcome !

Answer 17

can you help me with this one?

Answer 18

Sure

Answer 19

-5 + 2x^2 = -6x (what expression gives the solution?)

Answer 20

well, first add 6x to both sides

then order in a polynomial degree x^2 then x then the number (the constant)

Answer 21

\[\frac{ 2\pm \sqrt{4-(4)(6)(-5)} }{ 12 }\]

Answer 22

^^^^ that is A

Answer 23

you would get \(\color{goldenrod}{ 2x^2+6x-5=0 }\)

ooh, working fast on that one !

Answer 24

but look at my equation and then do the quadratic, b/c I think you go things wrong.

Answer 25

2 is not B. you got confused, b/c you didn't put the equation in order. See what I mean?

Answer 26

\[\frac{ -5\pm \sqrt{25-(4)(2)(6)} }{ -10 }\]

Answer 27

^^^^^^ that is B, I am having to type them one at a time b/c my computer is slow and lagging!!!

Answer 28

no, incorrect again.

HOLD ON! first put the equation in a degree order. you will see what I mean.

\(\color{blue}{ -5+2x^2=-6x }\)
\(\color{red}{ ~~~+6x~~~~~~+6x }\)

\(\color{blue}{ -5+2x^2+6x=0 }\) put this in a normal order

\(\color{blue}{ 2x^2+6x-5=0 }\)

\(\color{blue}{ a=2~~~~~~b=6~~~~~~c=-5 }\)

do it.

Answer 29

C. \[\frac{ -6\pm \sqrt{36-(4)(2)(-5)} }{ 4 }\]
D. \[\frac{ 6\pm \sqrt{36-(4)(2)(5)} }{ 4 }\]

Answer 30

ok

Answer 31

knowing your a b c looking a what I said just now (the blue and red stuff in a prev. reply ) which one do you think it is ?

Answer 32

so it is C?

Answer 33

Yup !

Answer 34

thanks!!! :D

Answer 35

I would give you a million medals if I could!!! lol!!! :D

Answer 36

So for the first question I had, was that quadratic formula?

Answer 37

\(\color{blue}{ Nice~~~~~to~~~~~hear~~~~~that~,~~~~~you~~~~~welcome~! }\)

Answer 38

For the first? well the first question either way was no solution. If you got no solution doing the quadratic formula, you would get no solution using any other method as well.

Answer 39

would this be quadratic formula 8x^2 - 13x + 3 = 0?

Answer 40

and ok!!

Answer 41

would you solve this one using the quad. form. ?

Answer 42

that;s your question?

Answer 43

yes! I guess so!!

Answer 44

b/c it wouldnt be factoring graphing or square roots!!! lol

Answer 45

yeah, it would be very difficult to complete the square. You CAN'T factor it (into integers, but you could into decimals, but you won't try to do that, b/c ik you aren't stupid)

yeah, you would prefer quad. form. as your best method (in this case) .

Answer 46

Thanks!!! I have just one more question!! :) Could you help me with it?

Answer 47

Yeah, shoot !

Answer 48

The perimeter of a rectangle is 54 cm. The area of the same rectangle is 176 cm^2. What are the dimensions of the rectangle?

Answer 49

ok, let width be "w" and length be "l"

\[P=2(w+l)\]
\[A=~w \times l\]

substitute what you know for P (Perimeter) and A (Area) and solve.

\(\color{blue}{ 54=2(w+l) } \)
\(\color{blue}{ 176=w \times l } \)

Answer 50

doesn't matter which one is l and which one is w. (That you won't find)


can you solve the system of equation I posted in blue ?

Answer 51

I can try!!

Answer 52

yeah, no I got an answer that was way off from my choices!!

Answer 53

Substitute from the first equation into the second.

\(\color{blue}{ 54 =2(w+l) } \) \(\color{black}{~~~->~~~} \) \(\color{blue}{ 27 =w+l } \) \(\color{black}{~~~->~~~} \) \(\color{blue}{ 27-w =l } \)

substitute \(\color{blue}{ 27-w } \) for \(\color{blue}{ l } \).


\(\color{blue}{ 176= w \times l } \)
\(\color{blue}{ 176= w \times (27-w) } \)
\(\color{blue}{ 176= 27w-w^2 } \)


can you solve \(\color{blue}{ 176= 27w-w^2 } \) ?

Answer 54

176(27) - 176^2?

Answer 55

11 and 16 that's what I got. for w

Answer 56

oh ok, thank you!!! I really need help in this part of Algebra, good thing my mom is coming home soon!!! Thanks!!!

Answer 57

Have a great day!!! :) bye!!

Answer 58

yeah, ...
\(\color{blue}{ 11\times16=176 ~~~~~~~~~~~:D } \)

Answer 59

yeah, bye !