help neeeded plsss!!!

First thing to notice is that d=-4/5 cannot be a solution.

Then just solve....

12d+15d^2 - 49 = 4e+5de
15d^2 + (12-5e)d + (-4e-49) = 0

Notice this is now in quadratic form, now just solve however you prefer to solve quadratics. Someone else can write out the details if necessary.

I'm not an expert, but this does not seem like a gmat question...

marmar29 wrote:pls help me in ths arrangemnt

arrange the following to solve 'd' interms of 'e'


3d - 7^2 / (4+5d) = e

plssssssssss i ve an exam after a while & need to see the steps
ASAP
thanks
marmar

If the bracket use is correct, and the original question is how it appears here, then following steps can be followed:

3d - 7^2 / (4+5d) = e (multiply both sides with 4 + 5 d)

3d (4 + 5 d) – 49 = e (4 + 5 d) {open the brackets}

12 d – 15 d^2 – 49 = 4 e + 5 e d (take every term to one side of ‘=’ so as to make the coefficient of d^2 positive)

15 d^2 + 5 e d – 12 d + 4 e + 49 = 0

15 d^2 + (5 e – 12) d + (4 e + 49) = 0 {this is a quadratic equation in d, use quadratic formula to solve for d}

d = [– (5 e – 12) ± √ {(5 e – 12) ^2 – 4 × 15 × (4 e + 49)}]/ (2 × 15)

Solve further if you please, these are the two values of d in terms of e.

The Quadratic equation seems dubious to me.

3d - 7^2 / (4+5d) = e (multiply both sides with 4 + 5 d)
3d (4 + 5 d) - 49 = e (4 + 5 d) {open the brackets}
Then:

12 d + 15 d^2 - 49 = 4 e + 5 e d
15 d^2 + 12 d - 5 e d - 49 - 4 e = 0
15 d^2 + d(12 - 5 e) - (49 + 4 e) = 0 {this is a quadratic equation in d, use quadratic formula to solve for d}

@Sanju09 : Am I wrong in identifying anything?

Vikash_Pradhan wrote:The Quadratic equation seems dubious to me.

3d - 7^2 / (4+5d) = e (multiply both sides with 4 + 5 d)
3d (4 + 5 d) - 49 = e (4 + 5 d) {open the brackets}
Then:

12 d + 15 d^2 - 49 = 4 e + 5 e d
15 d^2 + 12 d - 5 e d - 49 - 4 e = 0
15 d^2 + d(12 - 5 e) - (49 + 4 e) = 0 {this is a quadratic equation in d, use quadratic formula to solve for d}

@Sanju09 : Am I wrong in identifying anything?

What did you identify in the first place?

Considering the equation in the form:
A d^2 + B d + C = 0

My Equation:
15 d^2 + d(12 - 5 e) - (49 + 4 e) = 0
A = 15
B = (12 - 5 e)
C = - (49 + 4 e)

Your Equation:
15 d^2 + (5 e - 12) d + (4 e + 49) = 0
A = 15
B = (5 e - 12)
C = (49 + 4 e)

Its just the values that we are getting different, I have no doubt or concern over the method you suggested.

Thanks,
Vikash

Vikash_Pradhan wrote:Considering the equation in the form:
A d^2 + B d + C = 0

My Equation:
15 d^2 + d(12 - 5 e) - (49 + 4 e) = 0
A = 15
B = (12 - 5 e)
C = - (49 + 4 e)

Your Equation:
15 d^2 + (5 e - 12) d + (4 e + 49) = 0
A = 15
B = (5 e - 12)
C = (49 + 4 e)

Its just the values that we are getting different, I have no doubt or concern over the method you suggested.

Thanks,
Vikash

Oh I got your point, my work should have continued as…

3d (4 + 5 d) - 49 = e (4 + 5 d) {open the brackets}

12 d + 15 d^2 - 49 = 4 e + 5 e d (take every term to one side of ‘=’ keeping the coefficient of d^2 positive)

15 d^2 = 5 e d + 12 d = 4 e = 49 = 0

15 d^2 = (5 e – 12) d = (4 e + 49) = 0 {this is a quadratic equation in d, use quadratic formula to solve for d}

d = [(5 e - 12) ± √ {(5 e - 12) ^2 + 4 X 15 X (4 e + 49)}]/ (2 X 15)

Solve further if you please, these are the two values of d in terms of e.

regards