• Mar 3rd 2007, 02:35 PM
phatjigga
Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

Here goes,

For which real values of ‘a’ does the quadratic equation
F(x)=x^2-ax+12=0
Have its roots two consecutive integers? What are these roots?

-im not sure if this will help but I got ‘a’ as 7??—this was throught guess and check
• Mar 3rd 2007, 03:02 PM
Jhevon
Quote:

Originally Posted by phatjigga
Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

Here goes,

For which real values of ‘a’ does the quadratic equation
F(x)=x^2-ax+12=0
Have its roots two consecutive integers? What are these roots?

-im not sure if this will help but I got ‘a’ as 7??—this was throught guess and check

Yeah, this is a weird problem. It seems easy at first, but its weird. The following method is not one i'm proud of, but i dont see another way to do this. I tried completing the square, and the quadratic formula to no avail. But here's what i did.

x^2-ax+12=0

=> (x - m)(x - n) = 0 where m*n = 12 and m + n = a
the only two consecutive integers that multiply to give 12 is 3 and 4.
thus a = 3 + 4 = 7

The reason i dont like this method is because there is guessing involved at the end. with 12 it wasnt so bad, but if our number was a lot larger than 12, we might have had a rough time thinking of two consecutive integers to multiply and get it.

I'm sure another mathematician will come up with a better way to do this soon
• Mar 3rd 2007, 03:03 PM
topsquark
Quote:

Originally Posted by phatjigga
Hey I have another problem, I thought I could do this one but i think im just getting myself confused even more. Can you help me?

Here goes,

For which real values of ‘a’ does the quadratic equation
F(x)=x^2-ax+12=0
Have its roots two consecutive integers? What are these roots?

-im not sure if this will help but I got ‘a’ as 7??—this was throught guess and check

Well if the roots are to be integers then the discriminant has to be greater than or equal to 0. So:
(-a)^2 - 4*1*12 >= 0

a^2 >= 48

a >= sqrt(48) = 4*sqrt(3)

So two real roots are possible.

What are the roots of the equation in terms of a?
x = (a (+/-) sqrt(a^2 - 48))/2

If these two roots are to be consecutive integers then
x- = (a - sqrt(a^2 - 48))/2
x+ = (a + sqrt(a^2 - 48))/2
differ by 1.

So let (x+) = (x-) + 1

(a + sqrt(a^2 - 48))/2 = (a - sqrt(a^2 - 48))/2 + 1

a + sqrt(a^2 - 48) = a - sqrt(a^2 - 48) + 2

2*sqrt(a^2 - 48) = 2

sqrt(a^2 - 48) = 1

a^2 - 48 = 1

a^2 = 49

a = (+/-)7

Now, which one of these work? The easiest way is simply to solve the quadratics:
a = 7: x^2 - 7x + 12 = 0
Has solutions x = 3, 4. So a = 7 works.

a = -7: x^2 + 7x + 12 = 0
Has solutions x = -3, -4. So a = -7 also works.

-Dan
• Mar 3rd 2007, 03:06 PM
Jhevon
Quote:

Originally Posted by topsquark
Well if the roots are to be integers then the discriminant has to be greater than or equal to 0. So:
(-a)^2 - 4*1*12 >= 0

a^2 >= 48

a >= sqrt(48) = 4*sqrt(3)

So two real roots are possible.

What are the roots of the equation in terms of a?
x = (a (+/-) sqrt(a^2 - 48))/2

If these two roots are to be consecutive integers then
x- = (a - sqrt(a^2 - 48))/2
x+ = (a + sqrt(a^2 - 48))/2
differ by 1.

So let (x+) = (x-) + 1

(a + sqrt(a^2 - 48))/2 = (a - sqrt(a^2 - 48))/2 + 1

a + sqrt(a^2 - 48) = a - sqrt(a^2 - 48) + 2

2*sqrt(a^2 - 48) = 2

sqrt(a^2 - 48) = 1

a^2 - 48 = 1

a^2 = 49

a = (+/-)7

Now, which one of these work? The easiest way is simply to solve the quadratics:
a = 7: x^2 - 7x + 12 = 0
Has solutions x = 3, 4. So a = 7 works.

a = -7: x^2 + 7x + 12 = 0
Has solutions x = -3, -4. So a = -7 also works.

-Dan

See i told you, except it wasn't a mathematician, it was a physicist, but close enough. i did what topsquark did up to that "differ by 1" line. for some reason i missed that and so tried another way