Question

x^2-7x+5=0 Find the roots of the quadractic equation and obtain their sum and product.

Answer 1

\[x=\frac{-(-7) \pm \sqrt{(-7)^2-4(1)(5)}}{2(1)}\]

Answer 2

we can do the second question before the first

Answer 3

the product is 5, and the sum is 7
you can read that right off the the equation

Answer 4

so what if it a variable like ax^2+bx+c=0

Answer 5

\[x=\frac{7 \pm \sqrt{49-20}}{2}=\frac{7 \pm \sqrt{29}}{2}\]

Answer 6

\[\frac{7+\sqrt{29}}{2}+\frac{7-\sqrt{29}}{2}=\frac{14}{2}=7\]

Answer 7

@monkey131 i dont think they are looking for that

Answer 8

why?

Answer 9

they jusy wana fine the a+B and aB

Answer 10

you knew it was 7 in advance.
also you know the product is 5
if you have
\[x^2+bx+c=0\] and the roots are
\[r_1,r_2\] then
\[r_1\times r_2=c\] and
\[r_1+r_2=-b\]

Answer 11

\[\frac{7-\sqrt{29}}{2}\cdot \frac{7+\sqrt{29}}{2}=\frac{49-29}{2}=\frac{20}{10}=2\]

Answer 12

a + b = -b/a
ab = c/a

Answer 13

how'd u make it 49

Answer 14

@monkey

Answer 15

7 times 7

Answer 16

@Satellite : im still dont get it T.T

Answer 17

@monky denominator should be 4
product is 5

Answer 18

Which part of the question don't you get?

Answer 19

you are right satellite
so we have 20/4=5

Answer 20

@killer, if you have
\[x^2+bx+c=0\] and the roots are
\[r_1,r_2\] then it factors as
\[(x-r_1)(x-r_2)\]

Answer 21

If you complete the square of ax^2 + bx + c = 0 it would explain it better..I think

Answer 22

when you multiply out you get
\[x^2+bx+c=x^2-r_1x-r_2+r_1\times r_2\]
\[=x^2-(r_1+r_2)x+r_1r_2\]

Answer 23

therefore
\[r_1r_2=c\] and \[r_1+r_2=-b\]

Answer 24

ohh i see isee thank you :DD FINALY LOL

Answer 25

and nayedo is right, you will get the same thing if you use the quadratic formula and multiply out

Answer 26

shouldnt the sum be '-7'?

Answer 27

so lets use the quadratic equaltion i got to the part where 7+- (square root of 49-20)/2

Answer 28

a + b = \[(7 + \sqrt{29})/ 2 + (7-\sqrt{29})/ 2 = 14 / 2 = 7\]

Answer 29

\[ab = (7+\sqrt{29})/2 \times (7-\sqrt{29})/2 = (49 - 29)/4 = 5\]

Answer 30

@jahtoday it is
\[x^2+bx+c=x^2-(r_1+r_2)x+r_1r_2\]so
\[r_1+r_2=-b\]

Answer 31

yup u r correct