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# If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

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Joined: 02 Dec 2012
Posts: 173
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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17 Dec 2012, 09:12
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5% (low)

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82% (01:08) correct 18% (01:45) wrong based on 775 sessions

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If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2
Math Expert
Joined: 02 Sep 2009
Posts: 58453
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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17 Dec 2012, 09:14
4
1
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

$$\frac{4-x}{2+x}=x$$;

$$4-x=x*(2+x)$$;

$$4-x=2x+x^2$$;

$$x^2+3x-4=0$$.

As you can see there is no need to solve for x to answer the question.

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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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17 Dec 2012, 09:21
Simplify by keeping all the terms on side.
The equation becomes:

(4-x) = x*(2+x)

Now, expand the RHS of the equation:
(4-x) = 2x+x^2

Now, you see you already have a prt of the solution, a +ve x^2, so now, you need to move the LHS of the equation to the RHS and equate to 0:

0 = 2x+x^2-4+x

Simplify this further by adding 2x and x and voila!

x^2+3x-4 = 0

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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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23 May 2017, 09:04
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

$$\frac{(4 - x)}{(2 + x)} = x$$

Or, $$4 - x = 2x +x^2$$

Or, $$x^2 + 3x - 4 = 0$$

Thus, the answer must be (C) 0
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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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25 May 2017, 15:46
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

We can simplify the given equation by first multiplying each side of the equation by (2 + x).

(4 - x)/(2 + x) = x

4 - x = x(2 + x)

4 - x = x^2 + 2x

x^2 + 3x - 4 = 0

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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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12 Mar 2018, 10:28
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

guys by we already see from question that it is quadratic x^2 + 3x -4 and it always equals zero no ?
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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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12 Mar 2018, 11:59
dave13 wrote:
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

guys by we already see from question that it is quadratic x^2 + 3x -4 and it always equals zero no ?

Not quite.

On solving the given equation ,

$$(4-x)/(2+x) = x$$
$$4-x = x*(2+x)$$
$$4-x = x^2 + 2x$$
$$0 = x^2 + 3x - 4$$

Hence 0.
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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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13 Mar 2018, 00:56
dave13 wrote:
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

guys by we already see from question that it is quadratic x^2 + 3x -4 and it always equals zero no ?

Not quite.

On solving the given equation ,

$$(4-x)/(2+x) = x$$
$$4-x = x*(2+x)$$
$$4-x = x^2 + 2x$$
$$0 = x^2 + 3x - 4$$

Hence 0.

hey Gladiator59 thanks ! What do mean by "Not quite." do you mean that it is not always that quadratics equal ZERO ?
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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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13 Mar 2018, 01:10
1
dave13 wrote:

hey Gladiator59 thanks ! What do mean by "Not quite." do you mean that it is not always that quadratics equal ZERO ?

Yeah dave13,

Precisely so. "Quadratics" may or may not equal 0. What does always equal zero is a "Quadratic equation".

So for example.

$$x^2 +3x - 4 = 0$$ could be re-written as
$$x^2 + 3x -3 = 1$$ ( adding 1 to both RHS and LHS )

Here the LHS is a "quadratic" expression that equals 1 (RHS)

More generally speaking .. quadratic expressions are parabolas and where the intersect the x-axis they equal zero.

Best,
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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?  [#permalink]

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15 Sep 2019, 06:37
If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?

(A) -4
(B) -1
(C) 0
(D) 1
(E) 2

4-x = 2x + x^2
x^2+3x-4=0

IMO C

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Re: If (4 - x)/(2 + x) = x, what is the value of x^2 + 3x -4 ?   [#permalink] 15 Sep 2019, 06:37
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