Question

3x^2-2x+4=0
a. a Repeadted Real Solution
b. Two Unrequal Real Solution
c. No Real Solution

Answer 1

ok,we have the formula for this equation:

Answer 2

How do i tell the Difference?

Answer 3

the standard form of these equations is:
\(\sf \Large Ax ^{2} + Bx + C = 0\)

Answer 4

\(\sf \Large x _{1,2} = \frac{ -b \ \pm \ \sqrt{b ^{2}- \ 4ac} }{ 2a }\)

Answer 5

now can you do it yourself?

Answer 6

well i dont need the solution... just to know what they are..

Answer 7

like i know how to solve it no problem.. i dont know the diff between repeated or unequal or no real solution

Answer 8

first find x1 , x2 then i will tell you :)

Answer 9

may i write?

Answer 10

@oscargt1

Answer 11

sure

Answer 12

x1 = 1/3 (1+i sqrt(11)) x2 = 1/3 (1-i sqrt(11))

Answer 13

\(\sf \large x _{1,2} = \frac{ -(-2) \pm \sqrt{44} }{ 6 }\)

Answer 14

ok?

Answer 15

just don't calculate it,because the question wants us another thing ...

Answer 16

-1/6 (-2(+ or -)2 sqrt(11))

Answer 17

no,it's not necessary...when \(\sf \large\sqrt{b ^{2}-4ac}\) is equal to 0 we have one solution so a Repeadted Real Solution is right,but when it's equal to a negative number we don't have any solution so No Real Solution is right,and when it's qual to a positive number,we have two solutions so Two Unrequal Real Solution is right :)

Answer 18

got it?

Answer 19

Thanks so much!

Answer 20

you're welcome :)

Answer 21

oh sorry i made a mistake!fogot that!look at this:

Answer 22

no,it's not necessary...when \(\sf \LARGE b ^{2}-4ac\) is equal to 0 we have one solution so a Repeadted Real Solution is right,but when it's equal to a negative number we don't have any solution so No Real Solution is right,and when it's qual to a positive number,we have two solutions so Two Unrequal Real Solution is right :)

Answer 23

and do you want to know why?

Answer 24

because,when \(\sf b ^{2}-4ac = 0\) so we have \(\sf \sqrt{0}\) and \(\sf \sqrt{0} = 0\) so \( \sf -b (+ \ or \ -) 0 = -b\)

Answer 25

because,when \(\sf b ^{2}-4ac = 0\) so we have \(\sf \sqrt{0}\) and \(\sf \sqrt{0} = 0\) so \( \sf -b (+ \ or \ -) 0 = -b\)

Answer 26

but when \(\sf b^{2} \ - \ 4ac \) is equal to a positive number for example 4 we have \(\sf \sqrt{4}\) it s 2 so \(\sf -b + 2\) is a different to \( \sf -b - 2 \) so we have two solutions...

Answer 27

but when \(\sf b^{2} - 4ac \) is equal to a negative number for example -4 we don't have any solution be cause \(\sf \large \sqrt{-4}\) does not exist \(\sf (is \; undefined) \)

Answer 28

ok?