John and Jane started solving a quadratic equation. John mad

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.

Hi Brunel,

I've read through Viete's theorem and while it's quite handy, I'm still not sure when to use X1+X2=-b/a vs X1*X2=c/a. Do you mind clarifying an approach or a link to a page that clarifies Viete's theorem further?

Bumping for review and further discussion.

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.

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
@

Yes I would be counting on to the explaination of this one too.

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