Question

If the equation(a^2+b^2)x^2 - 2b(a+c)x + b^2+c^2 has equal roots then , which of the following is right and how ?
a) 2b=a+c b)b^2=ac c)b=2ac/a+c d)b=ac

Answer 1

a, b, c are constants?

Answer 2

must be

Answer 3

yeah !

Answer 4

you know about discriminants ?

Answer 5

yes . b^2-4ac !

Answer 6

\(b^2-4ac\) thingi ?

Answer 7

yeah, so just identify a,b,c here...

Answer 8

In a quadratic equation \(Ax^2+Bx+C=0\),
if \(B^2-4AC>0\), it has 2 solutions.
if \(B^2-4AC<0\), it has no real solutions.
if \(B^2-4AC=0\), it has only 1 solution.

Answer 9

i have learnt these rules !

Answer 10

cool :)

Answer 11

a = (a^2+b^2 b= - 2b(a+c) c= b^2+c^2

Answer 12

yep :)

Answer 13

correction :
a quadratic equation CANNOT have only one solution

Answer 14

but I would use capital letters A, B, C

Answer 15

when B^2-4AC = 0
we have 2 EQUAL solutions

Answer 16

to avoid ambiguity

Answer 17

ok !

Answer 18

@saravananr yep @hartnn is right sorry

Answer 19

you want to know where those rules came from ?

Answer 20

that's ok ! it's understandable !

Answer 21

i need the way to solve it ? i tried by discriminant method ! but couldn't come to the conclusion

Answer 22

\(\huge \dfrac{-b \pm \color{blue}{\sqrt{b^2-4ac}}}{2a}\)

Answer 23

quadratic formula !

Answer 24

So you need to set up an equation equating the discriminant and 0 :)

Answer 25

i did so !

Answer 26

yes, from the formula
think when we will have 2 equal solutions ?
when everything after \(\pm\) equals 0

Answer 27

-b/2a

Answer 28

lets pause this problem for some time and try to understand those rules ?
you can resume the problem any time u want....

Answer 29

ok !

Answer 30

because those rules are very important

Answer 31

ok

Answer 32

so you got how the rule came for equal solutions ?

Answer 33

i know ! i have learnt it

Answer 34

ohh, all rules ?

Answer 35

only major ones !

Answer 36

i mean, if you have any doubt in those rules, get them clarified first...

Answer 37

no , i do not have

Answer 38

good, then we can resume your problem,

Answer 39

best

Answer 40

so now you can set an equation that B^2-4AC=0 :)

Answer 41

or in other words, B^2=4AC

Answer 42

i am trying the same

Answer 43

[-2b(a+c)]^2=4(a^2+b^2)(b^2+c^2)

Answer 44

yeah

Answer 45

b^2(a+c)^2=(a^2+b^2)(b^2+c^2)

Answer 46

if i am not wrong , i have derived 2ab^2c-a^2c^2-b^4

Answer 47

i meant 2ab^2c-a^2c^2-b^4=0

Answer 48

\[\Large\begin{array}{rcl}
b^2(a+c)^2-(a^2+b^2)(b^2+c^2)&=&0\\
(a^2b^2+2ab^2c+b^2c^2)-(a^2b^2+a^2c^2+b^2c^2+b^4)&=&0\\
2ab^2c-a^2c^2-b^4&=&0\\
-1\times \color{blue}{b^4}+2ac\times \color{blue}{b^2}-a^2c^2&=&0
\end{array}\]

Answer 49

Notice that all options have \(b\) as the subject :)

Answer 50

ok, what u got was correct

Answer 51

yeah !

Answer 52

but what is the next step to get to the right ans ?

Answer 53

couldn't you notice that what u got was a perfect square trinomial?

Answer 54

like \(A^2+2AB+B^2\)

Answer 55

@hartnn wow even I didn't notice it until you reminded me lol

Answer 56

with -2AB insteead

Answer 57

like
\(A^2-2AB +B^2 = (A-B)^2\)

Answer 58

@hartnn I was like wow there comes a new quadratic equation but I didn't know it's perfect :)

Answer 59

so is it -(ac+b^2)^2 = 0 ?

Answer 60

2ab^2c-a^2c^2-b^4=0

(b^4 - 2(b^2)ac +(a^2c^2)) = 0

Answer 61

oh ! gr8 !

Answer 62

no...
(b^2 - ac)^2 =0

Answer 63

\[\Large\begin{array}{rcl}
b^2(a+c)^2-(a^2+b^2)(b^2+c^2)&=&0\\
(a^2b^2+2ab^2c+b^2c^2)-(a^2b^2+a^2c^2+b^2c^2+b^4)&=&0\\
2ab^2c-a^2c^2-b^4&=&0\\
-1\times \color{blue}{b^4}+2ac\times \color{blue}{b^2}-a^2c^2&=&0\\
b^4-2acb^2+a^2c^2&=&0\\
(ac-b^2)^2&=&0\\
ac-b^2&=&0\\
b^2&=&ac\\
b&=&\sqrt{ac}
\end{array}\]
All credits to @hartnn :D

Answer 64

got it very well ! Thanks lot to both of u !

Answer 65

no problem :)

Answer 66

both of u have helped me immensly . tomorrow is my exam . plz pray for me

Answer 67

wish you a very good luck! :)

Answer 68

thanks

Answer 69

gd luck :D

Answer 70

thanks

Answer 71

@kc_kennylau , and @hartnn r u there ?

Answer 72

i wan to share smth more

Answer 73

yes

Answer 74

sure, go on

Answer 75

i am student of class X living in KSA

Answer 76

i am an indian

Answer 77

im from hong kong, close to you :D

Answer 78

I have started an educational community for the students of both class IX and X where a lot many question papers revision notes and lot more are shared by various student from different parts of the world

Answer 79

meant for CBSE students

Answer 80

so , plz u can also contribute to the community by posting smth gud and educational

Answer 81

the link is iisjcommunity.co.nr

Answer 82

plz join and spread it to someone whom u all know too !

Answer 83

hope u will do this help ?

Answer 84

I just joined saravananr !!

Answer 85

thanks