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How many different values of x does the solution set for the equation

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Sajjad1994
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How many different values of x does the solution set for the equation [#permalink] New post   21 Mar 2020, 23:05
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59% (01:12) correct 41% (01:24) wrong based on 87 sessions
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How many different values of \(x\) does the solution set for the equation \(4x^2 = 4x - 1\) contain?

(A) None

(B) One

(C) Two

(D) Four

(E) Infinitely many
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B
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EgmatQuantExpert
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Re: How many different values of x does the solution set for the equation [#permalink] New post   22 Mar 2020, 07:20
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Solution



Given
In this question, we are given that
    • An equation \(4x^2 = 4x – 1\)

To find
We need to determine
    • The number of different values of x that satisfy the given equation

Approach and Working out
Solving the given equation, we get:
    • \(4x^2 = 4x – 1\)
    Or, \(4x^2 – 4x + 1 = 0\)
    Or, \(4x^2 – 2x – 2x + 1 = 0\)
    Or, \(2x (2x – 1) – 1 (2x – 1) = 0\)
    Or, \((2x – 1) (2x – 1) = 0\)
    Or, \((2x – 1)^2 = 0\)
    Hence, \(x = \frac{1}{2}\)


Thus, option B is the correct answer.

Correct Answer: Option B
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Re: How many different values of x does the solution set for the equation [#permalink] New post   25 Mar 2020, 10:32
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We can use the discriminant for solve this question.

Discriminant = (-4)^2 - 4 x(4)x(1) = 16 - 16 = 0
The equation has a unique solution.

Correct answer is B)
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Re: How many different values of x does the solution set for the equation [#permalink] New post   25 Mar 2020, 21:22
SajjadAhmad wrote:
How many different values of \(x\) does the solution set for the equation \(4x^2 = 4x - 1\) contain?

(A) None
(B) One
(C) Two
(D) Four
(E) Infinitely many


The equation can be rewritten as 0 = 4x^2 - 4x + 1. That equation can be factored into 0 = (2x-1)(2x-1). There is only one x value that can satisfy the equation and thus the answer is (B).
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Re: How many different values of x does the solution set for the equation [#permalink] New post   25 Mar 2020, 22:13
Rearrange the terms : 4x^2–4x+1=0

Upon looking closely we can see that the above equation is expansion of (2x–1)2=0
This equation has only one solution
x=1/2
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Re: How many different values of x does the solution set for the equation [#permalink] New post   09 Dec 2020, 02:42
Asked: How many different values of \(x\) does the solution set for the equation \(4x^2 = 4x - 1\) contain?

\(4xˆ2 - 4x + 1 = 0\)
\((2x - 1)ˆ2 = 0\)
x =0.5

IMO B
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Re: How many different values of x does the solution set for the equation [#permalink]
09 Dec 2020, 02:42
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