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John and Jane started solving a quadratic equation. John mad [#permalink]

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20 Dec 2010, 08:20

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A

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C

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E

Difficulty:

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20% (01:49) wrong based on 272 sessions

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John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

A. x^2 + 4x + 14 = 0 B. 2x^2 + 7x -24 = 0 C. x^2 -14x + 48 = 0 D. 3x^2 -17x + 52 = 0 E. 2x^2 + 4x + 14 = 0

John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0 b. 2x2 +7x -24 =0 c. x2 -14x +48 =0 d. 3x2 -17x +52 =0 e. 2x2 + 4x +14 =0

Kindly explain..!!

Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

John made a mistake while copying the constant term and got the roots as 5 and 9 so he copied coefficient of x and x^2 correctly: \(5+9=14=-\frac{b}{a}\);

Jane made a mistake in the coefficient of x and she got the roots as 12 and 4 so she copied the constant term and the coefficient of x^2 correctly: \(12*4=48=\frac{c}{a}\);

Only option C satisfies the above conditions \(x^2-14x+48=0\): \(-\frac{b}{a}=-\frac{-14}{1}=14\) and \(\frac{c}{a}=\frac{48}{1}=48\) (any of the condition would be sufficient).

John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0 b. 2x2 +7x -24 =0 c. x2 -14x +48 =0 d. 3x2 -17x +52 =0 e. 2x2 + 4x +14 =0

Kindly explain..!!

Viete's formula for the roots \(x_1\) and \(x_2\) of equation \(ax^2+bx+c=0\): \(x_1+x_2=-\frac{b}{a}\) AND \(x_1*x_2=\frac{c}{a}\).

John made a mistake while copying the constant term and got the roots as 5 and 9 so he copied coefficient of x and x^2 correctly: \(5+9=14=-\frac{b}{a}\);

Jane made a mistake in the coefficient of x and she got the roots as 12 and 4 so she copied the constant term and the coefficient of x^2 correctly: \(12*4=48=\frac{c}{a}\);

Only option C satisfies the above conditions \(x^2-14x+48=0\): \(-\frac{b}{a}=-\frac{-14}{1}=14\) and \(\frac{c}{a}=\frac{48}{1}=48\) (any of the condition would be sufficient).

Answer: C.

I just solve it with (x-y)(x-y) aproach not sure if I was just lucky and landed on the right answer or if it does always work

So: John got values 9 and 5 for x; (x-5)(x-9)=0 --> x^2-14x+45=0 (his mistake 45) Jane got values 12 and 4 for x; (x-12)(x-4)=0 --> x^2-16x+48=0 (her mistake 16x)

Eliminate mistakes and plug in the right values each picked: X^2-14x(from Jonhs equ)+48(from janes)=0

Guess that this might only work when X^2 does not have a multiplier...

Will be adding to my "Blackbook" Viete's formula for the roots.

John and Jane started solving a quadratic equation. John made a mistake while copying the constant term and got the roots as 5 and 9. Jane made a mistake in the coefficient of x and she got the roots as 12 and 4.What is the equation?

a. x2 + 4x +14 =0 b. 2x2 +7x -24 =0 c. x2 -14x +48 =0 d. 3x2 -17x +52 =0 e. 2x2 + 4x +14 =0

I just solve it with (x-y)(x-y) aproach not sure if I was just lucky and landed on the right answer or if it does always work

So: John got values 9 and 5 for x; (x-5)(x-9)=0 --> x^2-14x+45=0 (his mistake 45) Jane got values 12 and 4 for x; (x-12)(x-4)=0 --> x^2-16x+48=0 (her mistake 16x)

Eliminate mistakes and plug in the right values each picked: X^2-14x(from Jonhs equ)+48(from janes)=0

Guess that this might only work when X^2 does not have a multiplier...

Will be adding to my "Blackbook" Viete's formula for the roots.

This approach isn't incorrect. It is perfectly fine and Viete's formula is good to know (I assume it is discussed in detail in high school in most curriculums).

It will work even if the equation has a co-efficient other than 1 for x^2. Lets say, rather than x2 -14x +48 =0, the given equation in options is: 2x^2 - 28x +96 =0. It doesn't matter since 2 is common to all terms and will be taken out and eliminated. So, in essence, the equation is still x2 -14x +48 =0. Also if the equation is something like 4x^2 -28x + 45 =0, where nothing is common, the roots you obtain will reflect the co-efficient of x^2. i.e. roots of this equation are 5/2 and 9/2 and when you put it in the (x-a)(x-b) = 0 form, you get: (x - 5/2)(x - 9/2) = 0 x^2 - 14/2 x + 45/4 = 0 4x^2 - 28x + 45 = 0

So according to the given roots, you will always get the accurate equation. It might have something common in all the terms but that equation will still be the same.
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Re: John and Jane started solving a quadratic equation. John mad [#permalink]

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09 Oct 2014, 05:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: John and Jane started solving a quadratic equation. John mad [#permalink]

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09 Oct 2014, 07:30

1

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John made a mistake while copying the constant term and got the roots as 5 and 9. (x-5)(x-9)=0 x^2-14x+45=0 ....... I

Jane made a mistake in the coefficient of x and she got the roots as 12 and 4. (x-12)(x-4)=0 x^2-16x+48=0 ....... II

John made a mistake while copying constant term, so ignore constant term and pick first term of his solution x^2 Jane made a mistake in the coefficient of x, so ignore that term and pick the last term of her solution +48

Therefore correct equation must contain x^2 as its first term and +48 as its last term. only C satisfies the condition.

I found this very hard. Can you tell me what is the level of this question?

The question isn't very hard. You can easily get the values for the co-efficient of x and the constant term from the two equations as shown by itnas above. I would say this is around 600 level.
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Re: John and Jane started solving a quadratic equation. John mad [#permalink]

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13 Jan 2016, 09:30

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: John and Jane started solving a quadratic equation. John mad [#permalink]

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21 Jan 2017, 03:08

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: John and Jane started solving a quadratic equation. John mad [#permalink]

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21 Jan 2017, 13:47

A second degree equation is AX ^ 2 + BX + C = 0

The statement indicates that John does not have correct C, therefore everything else is correct and by the properties of the solutions, we can associate: (5 + 9) / A = -B Then 14 / A = -B

In addition the statement states that Jane does not have correct B, therefore everything else is correct and by the properties of the roots, we can associate (12 * 4) / A = C Then 48 / A = C

We look at the alternatives and the only one, which associates a single value for A, is alternative C, the only value of A is 1.

Correct alternative C @
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