VeritasPrepKarishma wrote:
vicksikand wrote:
In a quadratic equation of the form:
x^2 +bx+c = x^2 + d^2+2dx
can we directly equate coeff's on both sides and say
b=2d and c=d^2 (I think we can - any views?)
Yes, you can if you know that b, c and d are constants. Usually, an accepted assumption (but there may be exceptions).
Why ask this question?
I came across a d.s problem where I had to find c and the reduced equation was of the form:
bx + c= 2dx +d^2
1) d=3
2) b=6
now the conventional way of solving for c would be to re arrange:
c = (2d-b)x + d^2
For each value of x (b,d given) we will have a unique value of c. Hence Answer (E)
If you know that c is not an algebraic expression but a constant, the answer will not be (E), it will be (D).
The x term has to be 0, since c is a constant. So 2d=b. If we know d = 3, c =9.
Again, for statement (2), if b = 6, it will mean d = 3 and hence c = 9.
Point is, if b, c and d are constants, then b has to be equal to 2d and c has to be equal to d^2.
On the flip side , if we equate coefficients ( b=2d and c=d^2), we get:
1: d=3, c=d^2=9
2: b=6, b=2d => d=3, c=d^2 = 9
Answer (D)
I don't think the above is correct.
Question 1:
vicksikand wrote:
In a quadratic equation of the form:
x^2 +bx+c = x^2 + d^2+2dx
can we directly equate coeff's on both sides and say
b=2d and c=d^2 (I think we can - any views?)
First of all
x^2 will cancel out and we'll have:
bx+c=d^2+2dx. Now if we are told that this equation is true for all
x-es then yes:
c=d^2 and
b=2d. But if we are not told that then no, we can not say this. For example
bx+c=d^2+2dx can be
4x+2=2x+1, which means that
x=-\frac{1}{2},
b=4,
c=2,
d=1. Generally the solution for
x would be
x=\frac{d^2-c}{b-2d} and for different values of the constants you'll get different values of
x.
Question 2:
vicksikand wrote:
I came across a d.s problem where I had to find c and the reduced equation was of the form:
bx + c= 2dx +d^2
1) d=3
2) b=6
now the conventional way of solving for c would be to re arrange:
c = (2d-b)x + d^2
For each value of x (b,d given) we will have a unique value of c. Hence Answer (E)
On the flip side , if we equate coefficients ( b=2d and c=d^2), we get:
1: d=3, c=d^2=9
2: b=6, b=2d => d=3, c=d^2 = 9
Answer (D)
Are you talking about this problem:
If b, c, and d are constants and x^2 + bx + c = (x + d)^2 for all values of x, what is the value of c?(1) d = 3
(2) b = 6
Note that we are told that
x^2+bx+c=(x +d)^2 for all values of x.
So, as
x^2+bx+c=(x+d)^2 is true "for
ALL values of
x" then it must hold true for
x=0 too -->
0^2+b*0+c=(0+d)^2 -->
c=d^2.
Next, substitute
c=d^2 -->
x^2+bx+d^2=x^2+2dx+d^2 -->
bx=2dx --> again it must be true for
x=1 too -->
b=2d.
So we have:
c=d^2 and
b=2d. Question:
c=?(1)
d=3 --> as
c=d^2, then
c=9. Sufficient
(2)
b=6 --> as
b=2d then
d=3 --> as
c=d^2, then
c=9. Sufficient.
Answer: D.
BUT if we were not told that x^2+bx+c=(x+d)^2 is true "for ALL values of x" then the answer would be C, not D or E.First of all
x^2+bx+c=(x+d)^2=x^2+2dx+d^2 -->
bx+c=2dx+d^2.
(1)
d=3 -->
bx+c=2dx+d^2=6x+9 -->
c=x(6-b)+9. Not sufficient as if
x=0 than
c=9 but if
x=1 and
b=1 then
c=14.
(2)
b=6 -->
bx+c=6x+c=2dx+d^2. The same here. Not sufficient.
For (1)+(2) we would have
6x+c=6x+9 -->
c=9. Sufficient.
Answer: C.
So generally if b, c, d,and e are constants and we have bx+c=dx+e we can not say that b=d and c=e.Hope it helps.
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