Find K if the number of values of x that satisfy the equation

Question

Find K if the number of values of x that satisfy the equation  |x^2 - 16x + 55|=K  is 3

A 7
B 8
C 9
D 10
E 11

chetan2u · Answer

|x^2 - 16x + 55|=K 
 |x^2 - 16x + 64-9|=K 
 |(x-8)^2-3^2|=k 
 |(x-8-3)(x-8+3)|=k 
 |(x-11)(x-5)|=k

|(x-11)(x-5)|=|(12-11)(12-5)|=|1*7|=7 
 |(x-11)(x-5)|=|(4-11)(4-5)|=|-1*-7|=7 
So x can be 12 or 4

|(x-11)(x-5)|=|(7-11)(7-5)|=|-4*2|=8 
 |(x-11)(x-5)|=|(9-11)(9-5)|=|-2*4|=8 
So x can be 7 or 9

|(x-11)(x-5)|=|(8-11)(8-5)|=|3*3|=9 
So x can be 8.

How can k as 9 give 3 values of x??

revati16 · Answer

Can someone please elaborate on the answer a little. I am unable to understand this question

the27s · Answer

Quadratic has 2 solutions (can be distinct or identical), attach a modulus to it, the equation may have 2 more solutions (again distinct or identical).

As it is given that 3 solutions exist for x, one of the 2 quadratics (obtained by expanding the modulus) should have equal real roots.

Testing for b^2-4ac = 0, we get k=9/-9

That allows us to select option C.

But k is supposed to be positive, so this quadratic having 3 solutions seems incorrect.

Maybe the question should've read the equation satisfies 1 value of x.

Posted from my mobile device

Bunuel · Answer

It's not necessary x to be an integer, so for  k = 9 ,  |x^2 - 16x + 55|=9  has three solutions:
 x = 8 - 3 \sqrt {2} ;
 x = 8 + 3 \sqrt {2} ;
 x=8 .

If k is 7 or 8, the equation has four solutions, and if k is 10 or 11, the equation has two solutions.

adityashinde88 · Answer

Please can you explain how K=9?

khanabd8 · Answer

Case 1 :
x^2 - 16x + 55 - k = 0

Determinant :
(-16)^2 - 4(1)(55 - k)
= 256 - 220 + 4k
= 36 + 4k

Case 2 :
-x^2 + 16n - 55 - k = 0

Determinant :
(16)^2 - 4(-1)(-55 - k)
=256 - 220 - 4k=
= 36 - 4k

---

Determinant :
 
 (+ve) >0 has 2 solutions 
 = 0 has 1 solution 
 −ve <0 has no real solution (or 2 imaginary solutions not GMAT scope)

- Also : K is always (+ve) >0 as K = absolute Value expression

So for 3 solutions one of the determinants need to be = 0

Case 1 :
36+4k always (+ve) >0 as K always (+ve) >0
 Case 2 :
36-4k =0 (for 1 solution)
-4k = -36
4k = 36
k = 9

Choice: C

Kinshook · Answer

Find K if the number of values of x that satisfy the equation  |x^2 - 16x + 55|=K  is 3

|x^2 - 16x + 55| = |(x-11)(x-5)| = K

Either x^2-16x+55 = K 
x^2 - 16x + (55-K) = 0
Which will have real roots provided 16^2-4(55-K)>=0

Or x^2-16x+55 = -K 
x^2 - 16x + (55+K) = 0
Which will have real roots provided 16^2-4(55+K)>=0

If K=0
x^2-16x+55=(x-11)(x-5)=0; x=11 or 5; 2 roots; Not possible since the number of values of x that satisfy the equation is 3

Normally the equation combined will have 4 roots but they can have 3 roots if
1. One of the equation has equal roots
2. The equations have a common root

Case 1. One of the equation has equal roots
a. x^2 - 16x + (55+K) = 0 has equal roots
16^2 - 4(55+K) = 0
K = 9
b. x^2 - 16x + (55-K) = 0 has equal roots
16^2-4(55-K)=0
K = -9

Case 2. The equations have a common root
x will satisfy both equations
x^2 - 16x + (55+K) = 0
x^2 - 16x + (55-K) = 0
K = 0; 
x^2-16x+55=(x-11)(x-5)=0; x=11 or 5; 2 roots; Not possible since the number of values of x that satisfy the equation is 3

K = {9,-9}

Since K=-9 is not provided, K = 9

IMO C

QuantMadeEasy · Answer

To get three roots of the two given quadratic is possible when both the roots of an equation are equal.

Here's the detailed solution:
 https://www.youtube.com/watch?v=z5NVvSNaeAk

Smart Prep (Tutor)

sushanth21 · Answer

one case
x^2 - 16x + 55 = k
second case
x^2 - 16x + 55 = -k

roots for case 1 = (-16+-sqrt{16^2 - 4(55-k)})/2
roots for case 2 = (-16+-sqrt{16^2 - 4(55+k)})/2

To have only 3 roots, one of these under roots must become zero, as 4*64 = 256 which is equal to 16^2
we can make the under root in case 2 as 0, because we can get the value 64 only in the case 2 in (55+k)
16^2 - 4(55+k) = 0
k = 9

Find K if the number of values of x that satisfy the equation

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Find K if the number of values of x that satisfy the equation

Prep Toolkit

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