If k is a negative constant and x is a variable, what is the value of

Question

If k is a negative constant and x is a variable, what is the value of k if one of the roots of the equation k(x - 1)^2 = 5x - 7 is double the other?

A. -1
B. -2
C. -3
D. -5
E. -25

Kinshook · Answer

Asked: If k is a negative constant and x is a variable, what is the value of k if one of the roots of the equation k(x - 1)^2 = 5x - 7 is double the other?

k(x - 1)^2 = 5x - 7 
 k(x^2 - 2x + 1) = kx^2 - 2kx + k = 5x -7 
 kx^2 - (2k+5) + k+7 = 0

Let the roots of the equation be r and 2r

r + 2r = 3r = (2k + 5)/k;  
 r = (2k+5)/3k  (1) 
 2r^2 = (k+7)/k  
 r^2 = (k+7)/2k  (2)

(k+ 7)/2k = (2k+5)^2/9k^2 
 9k^2(k+7) = 2k(2k+5)^2 = 2k(4k^2 + 20k + 25) = 8k^3+40k+50 
 9k^3 + 63k^2 = 8k^3+40k+50k 
 k^3 + 23k^2 - 50k = 0 
k(k+25)(k-2)=0

k = -25; since k is a negative integer

IMO E

rajatchopra1994 · Answer

Explanation:
K(x - 1)^2 = 5x - 7 
Simplfying above Equation:
K(x^2 + 1 - 2x) = 5x - 7
Kx^2 - (2K+5)x + 7+K = 0

as given roots is double the other, let one root is r , 2nd will be 2r.

So sum of roots:
r+2r = 2K+5 / K
3r = 2K+5 / K 
r= 2K+5/ 3k --- 1

Product of roots:
2r x r = K+7 / K
2r^2 = K+7/K --- 2

From 1 & 2, Substitute r value from 1 in 2 ; we get
K^2 + 23k - 50 = 0
(K-2)(K+25) =0
K=2 , -25

Only -ve value, so K = -25

IMO-E

yashikaaggarwal · Answer

k(x - 1)^2 = 5x - 7
K(x^2+1-2x) = 5x-7
K(x^2)+k-2kx = 5x-7
K(x^2)+k+7-2kx-5x = 0
K(x^2)+(k+7)-x(2k+5) = 0
let Roots = a,2a
product of roots= k+7/k = 2a*a
k+7/k = 2(a^2) ---------(1)
sum of roots = 2k+5/k = 2a+a
2k+5/k = 3a 
or 2k+5/3k = a ----------(2)

putting value of a in equation 1
k+7/k = 2((2k+5/3k)^2)
=> k^2+23k-50 = 0
=> (k-2)(k+25)
=> K=2 and -25

since k is a negative constant K = -25

IMO E

ScottTargetTestPrep · Answer

Solution:

Simplifying the equation, we have:

k(x^2 - 2x + 1) = 5x - 7

kx^2 - 2kx + k = 5x - 7

kx^2 - (2k + 5)x + k + 7 = 0

Now, let’s check the given values in the answer choices:

A. -1

-x^2 - 3x + 6 = 0
x^2 + 3x - 6 = 0

This is not factorable. We eliminate A.

B. -2

-2x^2 - x + 5 = 0
2x^2 + x - 5 = 0

This is not factorable. We eliminate B.

C. -3

-3x^2 + x + 4 = 0
3x^2 - x - 4 = 0
(3x - 4)(x + 1) = 0

3x - 4 = 0 → x = 4/3

x + 1 = 0 → x = -1

However, 4/3 is not twice -1, so k can’t be -3. We eliminate C.

D. -5

-5x^2 + 5x + 2 = 0
5x^2 - 5x - 2 = 0

This is not factorable. We eliminate D.

By default, E must be the correct answer, but let’s verify that it’s correct. .

E. -25

-25x^2 + 45x - 18 = 0
25x^2 - 45x +18 = 0
(5x - 3)(5x - 6) = 0

5x - 3 = 0 → x = 3/5

5x - 6 = 0 → x = 6/5

We see that 6/5 is twice 3/5, so k = -25.

Answer: E

hiranmay · Answer

If k is a negative constant and x is a variable, what is the value of k if one of the roots of the equation k(x - 1)^2 = 5x - 7 is double the other?

A. -1
B. -2
C. -3
D. -5
E. -25-->   correct : k(x - 1)^2 = 5x - 7 => x^2-2x+1=5x/k-7/k => x^2-(2+5/k)x+(1+7/k)=0 --(i), Now one of the roots of the equation is  a , so other one is  2a , (x-a)(x-2a)=0=> x^2-3ax+2a^2 =0 --(ii), now (i) & (ii) are same equation, so 1+7/k=2a^2>=0 =>7/k>-1 => k<-7 and only answer which is less than -7 is -25

ouslam · Answer

we can find the answer quicker if we check the equation  r^2 = (k+7)/2k  , k needs to be positive or smaller than  -7  (because  r^2  needs to be positive) , the only value that does that among the choices is -25 , hence E

Rogermoris · Answer

K(x - 1)^2 = 5x - 7 
Simplfying above Equation:
 Kx^2 - (2K+5)x + 7+K = 0

We know that the roots is double the other, assuming one to be 'a' and other to be '2a'
Sum of roots:
 a+2a = \frac{(2K+5)}{K} 
 3a = \frac{2K+5}{K} 
 a= \frac{2K+5}{3k}  - (i)

Product of roots:
 2a*a = \frac{K+7}{K} 
 2a^2 = \frac{K+7}{K}  - (ii)

Using the above two equations, we solve for k
 K^2 + 23k - 50 = 0 
 (K-2)(K+25) =0 
 K=2 , -25

K is defined to be negative, so  K = -25

OPTION

erwr43534 · Answer

Why is 2k+5 divided by k? Similarly, why is k+7 divided by 2k?

bumpbot · Answer

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If k is a negative constant and x is a variable, what is the value of

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