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John and Jane started solving a quadratic equation. John mad

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

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New post Updated on: 09 Jul 2013, 10:02
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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

Originally posted by krishnasty on 20 Dec 2010, 08:20.
Last edited by Bunuel on 09 Jul 2013, 10:02, edited 1 time in total.
Edited the question.
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Re: Problem in the roots  [#permalink]

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New post 20 Dec 2010, 08:41
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krishnasty wrote:
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.
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Re: Problem in the roots  [#permalink]

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New post 21 Dec 2010, 04:56
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Bunuel wrote:
krishnasty wrote:
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 :lol:

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.
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Re: Problem in the roots  [#permalink]

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New post 21 Dec 2010, 13:43
itnas wrote:
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 :lol:

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

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New post 09 Oct 2014, 07:30
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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.

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

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New post 23 Dec 2014, 15:15
I found this very hard. Can you tell me what is the level of this question?
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Re: John and Jane started solving a quadratic equation. John mad  [#permalink]

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New post 23 Dec 2014, 21:35
Salvetor wrote:
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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New post 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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Re: John and Jane started solving a quadratic equation. John mad  [#permalink]

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Re: John and Jane started solving a quadratic equation. John mad   [#permalink] 30 May 2018, 09:04
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