Question

equation (a²+b²)x²-2(ac+bd)x+(c²+d²)=0 are equal,
prove that a/b=c/d.

Answer 1

Using the Quadratic Formula, we obtain:
\[x=\frac{2(ac+bd) \pm \sqrt{(-2(ac+bd))^2-4(a^2+b^2)(c^2+d^2)}}{2(a^2+b^2)}\]
but I'm not sure how we can show a/b=c/d from this

Answer 2

why did you say are equal?
that's what the = sign signs say

Answer 3

so if we can show ad-bc=0 we can show a/b=c/d
so if we can ad-bc is a zero then we win

Answer 4

bca mathematics this ,how to solve i don't yar plz help me

Answer 5

i was just asking because i thought you may be missing something since
you have=0 are equal
simplify that that i have above

Answer 6

\[x=\frac{2(ac+bd) \pm \sqrt{4(ac+bd)^2-4(a^2c^2+a^2d^2+c^2b^2+b^2d^2)}}{2(a^2+b^2)}\]
\[x=\frac{2(ac+bd) \pm \sqrt{4} \sqrt{(ac+bd)^2-(a^2c^2+a^2d^2+c^2b^2+b^2d^2)}}{2(a^2+b^2)}\]
\[x=\frac{2(ac+bd) \pm 2 \sqrt{(ac+bd)(ac+bd)-a^2c^2-a^2d^2-c^2b^2-b^2d^2}}{2(a^2+b^2)}\]
\[x=2 \cdot \frac{(ac+bd) \pm \sqrt{a^2c^2+acbd+acbd+b^2d^2-a^2c^2-a^2d^2-c^2b^2-b^2d^2}}{2(a^2+b^2)}\]
\[x=\frac{ac+bd \pm \sqrt{2acbd-a^2d^2-c^2b^2}}{a^2+b^2}\]
\[x=\frac{ac+bd \pm \sqrt{-(a^2d^2-2acbd+c^2b^2)}}{a^2+b^2}\]
\[x=\frac{ac+bd \pm \sqrt{-(ad-cb)^2}}{a^2+b^2}\]
\[x=\frac{ac+bd \pm \sqrt{(ad-cb)^2} \sqrt{-1}}{a^2+b^2}\]
\[x=\frac{ac+bd \pm (ad-cb)i}{a^2+b^2}\]
....

Answer 7

yeah i don't think we can show a/b=c/d unless your teacher said such that the equation only has real root(s)

Answer 8

if she said that part then ad-cb would have to =0 in order for us to
have ad-cb=0
ad=cb

a/b=c/d

Answer 9

right. If the roots of the equation are equal, then ....

Answer 10

i was trying to figure out why he said =0 are equal

Answer 11

like there is suppose to be something there?

Answer 12

definitely

Answer 13

So if we wrote this equation as
Ax^2 + Bx + C = 0
the roots are equal if
B^2 - 4AC = 0

Now
A = a^2 + b^2
etc.

Answer 14

Then substitute and grind away a little

Answer 15

right
is that what you think he means by are equal?

Answer 16

yes, definitely

Answer 17

ok
i was having problem with why would someone say something is equal twice?
a=b are equal lol

Answer 18

= means equal so why you say a=b are =
its just weird

Answer 19

thanks James

Answer 20

So, let's just check for a minute:
A = a²+b²
B = -2(ac+bd)
C = c²+d²

Then B^2 - 4AC
= 4[ (ac+bd)^2 - (a^2 + b^2)(c^2 + d^2) ]
= 0

if and only if

2abcd = (ad)^2 + (bc)^2

i.e.,

(ad - bc)^2 = 0

i.e., a/b = c/d