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.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:

PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?

25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership