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# If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -

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If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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09 Apr 2018, 10:28
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Difficulty:

35% (medium)

Question Stats:

70% (01:15) correct 30% (01:30) wrong based on 71 sessions

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If $$2 x^2 - 2x - 12 = 0$$ and $$y^2 - 5y + 6 = 0$$ when $$x = -y$$, then what is the value of $$x$$ ?

a. -3
b. -2
c. 0
d. 2
e. 3

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If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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09 Apr 2018, 14:37
QZ wrote:
If $$2 x^2 - 2x - 12 = 0$$ and $$y^2 - 5y + 6 = 0$$ when $$x = -y$$, then what is the value of $$x$$ ?

a. -3
b. -2
c. 0
d. 2
e. 3

• Simplify and factor the first quadratic equation
$$2 x^2 - 2x - 12 = 0$$
$$x^2 - x - 6 = 0$$
$$(x - 3)(x + 2) = 0$$

$$x = 3$$ OR
$$x = -2$$

• Factor the second quadratic equation
$$y^2 - 5y + 6 = 0$$
$$(y - 2)(y - 3) = 0$$

$$y = 2$$ OR
$$y = 3$$

• Heed the condition:
when $$x = -y$$

For values of x and y, only one pair yields

$$x = -y$$
when
$$x = -2$$and $$y = 2$$

When $$x = -y$$:

$$x = -2$$

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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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09 Apr 2018, 21:31
1
QZ wrote:
If $$2 x^2 - 2x - 12 = 0$$ and $$y^2 - 5y + 6 = 0$$ when $$x = -y$$, then what is the value of $$x$$ ?

a. -3
b. -2
c. 0
d. 2
e. 3

We can get the value of x without solving quadratics.

Substitute y = -x into $$y^2 - 5y + 6 = 0$$: $$x^2 + 5x + 6 = 0$$.

Multiply by 2: $$2x^2 + 10x + 12 = 0$$.

Subtract the first equation: $$2x^2 + 10x + 12 -(2 x^2 - 2x - 12)= 0$$;

$$12x+24=0$$;

$$x=-2$$.

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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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10 Apr 2018, 04:22
Ans B, solving for Y it gives value for Y = 2 or 3. Using equation X = -Y and solving equation 2 X^2 - 2X - 12 = 0 for X = -2, satisfies the equation hence Ans B

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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = -  [#permalink]

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11 Apr 2018, 16:19
QZ wrote:
If $$2 x^2 - 2x - 12 = 0$$ and $$y^2 - 5y + 6 = 0$$ when $$x = -y$$, then what is the value of $$x$$ ?

a. -3
b. -2
c. 0
d. 2
e. 3

Dividing the first equation by 2 and solving for x, we have:

x^2 - x - 6 = 0

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

x = 3 or x = -2

Solving for y in the second equation, we have:

(y - 3)(y - 2) = 0

y = 3 or y = 2

Since x = -y, we see that x must be -2 (and y must be 2).
Alternate Solution:

Note that 2x^2 can also be expressed as 2x^2 = 2x + 12 using the first equation.

Since x = -y, let’s substitute y = -x in the second equation:

(-x)^2 -5(-x) + 6 = 0

x^2 + 5x + 6 = 0

Let’s multiply each side of this equality by 2:

2x^2 + 10x + 12 = 0

2x^2 = -10x - 12

Since we have two expressions for 2x^2, let’s set them equal to each other:

2x + 12 = -10x -12

12x = -24

x = -2

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Re: If 2 x^2 - 2x - 12 = 0 and y^2 -5y + 6 = 0 when x = - &nbs [#permalink] 11 Apr 2018, 16:19
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