abhishekswamy wrote:1.)If the roots of the equation px^2+rx+r=0 are in the ratio a:b, then the value of sqrt{b/a}+sqrt{a/b} is
A. sqrt{1/r}
B. sqrt{r/p}
C. sqrt{l/p}
D. sqrt{1/p}
E. sqrt{3/p}
2.) If 3/(x+3)+ 4/(x+4)=2/(x+2)+5/(x+5), then the value of x is?
[spoiler]
source:
https://gmat.jumbotests.com/tests/gmat-q ... -questions[/spoiler]
Question 1.
Given: px^2+rx+r=0 , and
(root 1)/(root 2) =r1/r2 = a/b
Required: Sqrt(b/a)+ Sqrt(a/b)
= sqrt(r2/r1) + sqrt(r1/r2)
= (r2 + r1)/sqrt(r1*r2)
In a quadratic equations, the
sum of the roots = -b/a whereas, the
product of the roots = c/a where a is the coefficient of x^2, b is the coefficient of x and c is the constant.
Therefore, = (-r/p)/ sqrt. (r/p)
To simplify the process of finding the answer, let us represent
r/p by 'a'.
Hence, = -a/sqrt(a)
Now rationalize the denominator by multiplying the function by sqrt(3)/sqrt(3). That should give you: [-a * sqrt(a)]/[a]
= - [sqrt(a)]
Now let us get back to where we've introduced a. There, we replaced r/p by a; so it's time now to reverse that action.
- [sqrt(a)]=
-[sqrt(r/p)]. But this answer is not among the choices.Maybe when writing choice B, you missed the '-'sign.
Question 2.
3/(x+3) + 4/(x+4) = 2/(x+2) + 5/(x+5)
Plugging values from the choices is a better approach than the one I try to demonstrate.
Ok! Let's start:
To simplify our work, let's first represent X+3 by 'a'.
Therefore, we have: 3/(a) + 4/(a+1) = 2/(a-1)+ 5/(a+2)
Solving for 'a' shall give you: a=3 or a=-1/2
Now,find the value of x.
As we've represented X + 3 = a
X + 3 = 3 or X + 3 = -1/2
X= 0 or
X= -7/2
