Question

use the quadratic formula to solve the equation -x^2+6x-5=0
A. -5,-1
B. 1,5
C. -5,11
D. 2,10

Answer 1

a=-2, b= +6 , c=-5
by using quadratic formula
\[\frac{ -b \pm \sqrt{b^{2}-4ac} }{ 2a }\]
put the values

\[\frac{ -6\pm \sqrt{-6^{2}-4(-2)(-5)} }{ 2(-2) }\]
\[\frac{ -6\pm \sqrt{36-40} }{ -4 }\]
\[\frac{ -6\pm \sqrt{-4} }{ -4 }\]
\[\frac{ -6\pm 2i }{ -4 }\]

Answer 2

-6+2i\-4 OR -6-2i\-4
-2taking common from both equations
\[\frac{ -2(3-i) }{ -4 } OR \frac{ -2(3+i) }{ -4 }\]
\[\frac{ 3-i }{ 2 } OR \frac{ 3+i }{ 2 }\]
this is the answer

Answer 3

Haseeb96, a= -1 (not -2)
so your answer is not quite right

Answer 4

@rockstargirl17 ... hust use the equation hasseeb posted, but ust the correct values for a b and c
y = -x^2+6x-5 = 0
ax^2 + bx + c

so a = -1
b = 6
c = -5

Answer 5

*just use ... hust isn't a word ... yet

Answer 6

where do you plug it in to though im so confused

Answer 7

quadratic formula:
0 = -b + or - sqrt (b^2 - 4*a*c) / 2a

so looks like:\[0 = \frac {-b \pm \sqrt (b^2 - 4 \times a \times c)}{ 2a}\]

Answer 8

so it would be 0=-6+squ(6^2-4x-1x-5/ 2(-1)
right?

Answer 9

yep, that's a plus or minus sign tho, so you need to do both
so
0=-6+squ(6^2-4x-1x-5/ 2(-1)

or

0=-6-squ(6^2-4x-1x-5/ 2(-1)

and your using the x as times... right? (not as X)

Answer 10

yes im using it as times

Answer 11

sweet, so work out both equations, and you'll get 2 different values. those values are x values for when y = 0
those are your answers

Answer 12

so like do you just do the top first or the bottom and how would you get to answers

Answer 13

We have,

\(\displaystyle -x^2+6x-5=0\Leftrightarrow-ax^2+bx-c=0\)

\(a=-1\\
b=6\\
c=-5\)

The quadratic formula is,

\(\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

So therefore we have,

\(\displaystyle x=\frac{-6\pm\sqrt{6^2-4(-1)(-5)}}{2(-1)}=\frac{-6\pm\sqrt{36-20}}{-2}=\frac{-6\pm\sqrt{16}}{-2}=\)

\(\displaystyle x=\frac{-6\pm4}{-2} \)

We should get 2 values for x.

Answer 14

so it's like:
y = 0 @ x = -6+squ(6^2-4x-1x-5/ 2(-1)
= -6 + squ(36 - 20)/ -2
= -6 + sqrt 16 / -2
= -6 + 4 / -2
= -2/-2
= 1

so y = 0 @ x = 1 that's one point (1, 0)

now find the other:
y = 0 @ x = -6 - squ(6^2-4x-1x-5/ 2(-1) (note the minus sign)

Answer 15

so can u solve the 2nd x value @rockstargirl17 ?

Answer 16

If my notation was confusing...

\(\displaystyle x=\frac{-6\pm4}{-2}\)

\(\displaystyle~~\rightarrow\frac{-6+4}{-2}=\frac{-2}{-2}=1\)

\(\displaystyle~~\rightarrow\frac{-6-4}{-2}\frac{-10}{-2}=~?\)

Answer 17

okay so all together i got sq(160)/-2
12.64/-2 =
i dont think im doing it right

Answer 18

hmm... should be:
y = 0 @ x = { -6 - squ(6^2 - 4x-1x-5 }/ 2(-1)
= { -6 - squ(36 - 20) }/ -2
= {-6 - sqrt 16} / -2
= {-6 - 4} / -2
= -10/-2
x = 5

so 2nd point is y = 0 at x = 5 (5, 0)

Answer 19

@rockstargirl17
did you look at the work that I posted up above. That should tell you where you went wrong in your application of the quadratic formula. Compare your work to mine.

Answer 20

not sure where u were getting the 160 from, sorry
@austinL 's instructions are a bit clearer if that helps?

Answer 21

so its 5,1 but wouldnt it be -5,-1 ?

Answer 22

nope, negative divided by a negative = positive

Answer 23

yeah but the answer choices say -5,-1 and i got the same thing

Answer 24

Not sure how they got that....

\(\displaystyle \frac{-a}{-a}=1\)

\(\displaystyle \frac{-a}{-b}=\frac{a}{b}\)