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

This is a quadratic eqn in a, so to have a unique value, the descriminant should be 0

=> (-2c)^2 - 4*2*(c^2-4) = 0
=> 4c ^2 = 32
=> c = 2sqrt2 or -2sqrt2

As per equation 1:

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.

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...........

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

Thanks Anshu, It was very informative & Helpful

anshumishra wrote:

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

hold on a minute. i still dont understand....
i found the answer is D instead of E and i read your solution and i dont understand, can you please explain it again?

diebeatsthegmat wrote:

anshumishra wrote:

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

hold on a minute. i still dont understand....
i found the answer is D instead of E and i read your solution and i dont understand, can you please explain it again?

diebeatsthegmat,

If you have got the answer D (which I guess is the OA), please post the solution. I might have made some mistakes, as I mentioned in my previous post. The idea was to understand how to solve and learn few things along the way ?

If you have any doubts, please post as goyalsau has posted. Hope you understand, may be if I don't get time, someone else will answer it.

Can't believe I have been using the word "descriminant" in place of "determinant" for so long. Sorry about that, It should be determinant.

Thanks[/u]