goyalsau wrote:anshumishra wrote:combining the two given equations :
(a-c)^2 + b^2 = 4
=> a^2 + c^2 - 2ac + b^2 = 4
=> 2a^2 - 2c*a +(c^2 -4) = 0 ----- 1
Till here i able to understand everything,
But what is the reasoning behind this statement
anshumishra wrote:
This is a quadratic eqn in a, so to have a unique value, the descriminant should be 0
I think here you solve the equation with the formula for equation
ax^2 +bx + c = 0 ,
-b (-ve , +ve ) sqrt ( b^2 - 4ac ) / 2a
But you neglected -b/2a why is that so??????
anshumishra wrote:
=> (-2c)^2 - 4*2*(c^2-4) = 0
=> 4c ^2 = 32
=> c = 2sqrt2 or -2sqrt2
From where you got this a = -2c/4 ????????
anshumishra wrote:
a = -2c/4
The question is interested in +ve value of a (a>0)
Hence, we need the -ve value of c , i.e ; -2sqrt2.
D.
I think i have asked too many questions........ But You can treat me as a novice student when it comes to algebra...........
No problem goyalsau,
You already know quite a lot, so should be easy to digest. Also, I probably found a typo in my earlier post
I think here you solve the equation with the formula for equation
ax^2 +bx + c = 0 ,
-b (-ve , +ve ) sqrt ( b^2 - 4ac ) / 2a
But you neglected -b/2a why is that so??????
You are right, I have used the formula you mentioned,
So, for any quadratic equation : ax^2+bx+ c =0, The two roots are :
[-b+sqrt ( b^2 - 4ac )]/2a and [-b-sqrt ( b^2 - 4ac )]/2a
sqrt ( b^2 - 4ac ) is called descriminant and is generally denoted using "D"
So, the two roots are :
[-b+D]/2a and [-b-D]/2a
The question asks us for a single value of "x" ; The only way to have is when D=0
Then both the roots become : -b/2a
From where you got this a = -2c/4 ????????
I derived the equation :
2a^2 - 2c*a +(c^2 -4) = 0 ----- 1
Compare it with;
Ax^2 + Bx + C = 0
A =2
B = -2c
C = c^2 -4
So, the unique root = -B/2A = 2c/4 [
I missed a -ve sign in my previous post]
or, a = 2c/4
Now, for a to be +ve, c should be +ve, hence c = 2sqrt2
It should be
E (Hopefully I am not making any mistake).
Let me know if you have any doubts left.
Thanks
