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

Question

If k is a positive 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

Determine 'k' if one of the roots of the equation k(x-1)^2 = 5x-7 is double the other.

a)-2
b)2
c)25
d)both a and b
e) :angry

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 = 2

IMO-B

Kinshook · Answer

Asked: If k is a positive 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) = 5x - 7
kx^2 - 2kx + k = 5x -7
kx^2 - (2k + 5)x + (k+7) = 0

Let the roots be a and 2a

3a = \frac{(2k+5)}{k} ; a = \frac{(2k+5)}{3k} 
 2a^2 = \frac{(k+7)}{k} ; a^2 = \frac{(k+7)}{2k}

\frac{(k+7)}{2k} = \frac{(2k+5)^2}{9k^2} 
 (k+7)9k^2 = (2k+5)^2*2k 
 9k^3 + 63k^2 = (4k^2 + 20k + 25)2k = 8k^3 + 40k^2 + 50k 
 k^3 + 23k^2 - 50k = 0 
k = 0
 k^2 + 25k - 2k - 50 = 0 
(k+25)(k-2) = 0

k = {0,-25,2}
Since k is a positive constant, k = 2

IMO B

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 positive constant K = +2

IMO B

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 - 7x + 8 = 0

This is not factorable. So we can skip to B.

B. 2

2x^2 - 9x + 9 = 0
(2x - 3)(x - 3) = 0

2x - 3 = 0 → x = 3/2

x - 3 = 0 → x = 3

We see that 3 is twice 3/2, so k = 2.

Answer: B

rcgard · Answer

I attempted to solve it using the following approach:

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

I then took product of roots as:
 \frac{2}{k} = 2(x-1)^2

and

Sum of roots as:
 \frac{5}{k}  = 3(x-1)

However I am faulting somewhere along the line here. Can this be problem solved using this approach some how?

Thanks!

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

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