- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,917

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,917

- Moderator
- #2

- Feb 7, 2012

- 2,792

A similar calculation using $f_2(x)$ instead of $f_1(x)$ confirms the solution $a=30$.

I hope I am not missing something here. The question asks for the largest value of $a$, but as far as I can see there is only the one possible value for $a$.

Last edited:

- Thread starter
- Admin
- #3

- Feb 14, 2012

- 3,917

Thanks for participating,If the quadratic polynomials $f_1(x)$ and $f_2(x)$ both divide the cubic polynomial $g(x)$ then they must have a factor in common. Any common factor of $f_1(x)$ and $f_2(x)$ must also be a factor of $2f_1(x) - f_2(x) = -15\bigl(x+\frac15a\bigr)$. Therefore $f_1(x) = x^2+(a-29)x-a = \bigl(x+\frac15a\bigr)(x-5)$ (the second factor has to be $x-5$ in order to make the constant term equal to $a$). Compare the coefficients of $x$ to see that $a-29 = \frac15a - 5$, from which $a=30$.

A similar calculation using $f_2(x)$ instead of $f_1(x)$ confirms the solution $a=30$.

I hope I am not missing something here. The question asks for the largest value of $a$, but as far as I can see there is only the one possible value for $a$.

Hmm, I can't see anything incomplete in your solution but I got two values of $a$, one is 0 and the other one is 30 and hence the largest $a$ would be 30.

Edit: I didn't realize you have already edited your post mentioned that you have found the missing case.

BTW, here is my solution:

Let $k$ be the common root of the two quadratic equations $f_1(x)=x^2+(a-29)x-a$ and $f_2(x)=2x^2+(2a-43)x+a$ to the polynomial $g(x)$.

By substituting $x=k$ into these two quadratic equations we get:

$f_1(k)=k^2+(a-29)k-a=0$ | $f_2(k)=2k^2+(2a-43)k+a=0$ |

$k^2+(a-29)k-a=0$ | $2k^2+(2a-43)k+a=0$---(2) |

$2k^2+2(a-29)k-2a=0$---(1) |

Now, subtract equation (2) from equation (3) and solve for $a$, we get

$(2k^2+2(a-29)k-2a)-(2k^2+(2a-43)k+a)=0$

$a=-5k$

Back substitute $x=k$ and $a=-5k$ into the quadratic equation of $f_1$ and $f_2$ respectively gives

$f_1(k)=k^2+(-5k-29)k+5k=0$ | $f_2(k)=2k^2+(2(-5k)-43)k-5k=0$ |

$k^2-5k^2-29k+5k=0$ $-4k^2-24k=0$ $k(k+6)=0$ $k=0$ or $k=-6$ | $2k^2-10k^2-43k-5k=0$ $-8k^2-48k=0$ $k(k+6)=0$ $k=0$ or $k=-6$ |

So, the largest value of $a$ must go with the smallest negative value of $k$, i.e. $a_{\text{largest}}=-5(-6)=30$.

Last edited:

- Mar 31, 2013

- 1,346

let $f1(x) = (x-m)(x-p)$

and let $f2(x) = 2(x-m)(x-q)$

comparing constant term

of f1(x) = mp = - a and 2 mq = a we get p = - 2q or m = 0 => a = 0

then taking the product and comparing coefficient of x

we get m+p = 29-a ...(1)

m – q = (43-2a)/2

or 2m – p = 43 – 2a ... (2)

solving (1) and (2) 3 m = (72-3a) or m = 24 – a

so p = 2m + 2a – 43 = 48 – 43 = 5

now - a = mp = 5(24-a) or 4a = 120 or a= 30 hence a = 30 is largest

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,917

Thanks for participating,

let $f1(x) = (x-m)(x-p)$

and let $f2(x) = 2(x-m)(x-q)$

comparing constant term

of f1(x) = mp = - a and 2 mq = a we get p = - 2q or m = 0 => a = 0

then taking the product and comparing coefficient of x

we get m+p = 29-a ...(1)

m – q = (43-2a)/2

or 2m – p = 43 – 2a ... (2)

solving (1) and (2) 3 m = (72-3a) or m = 24 – a

so p = 2m + 2a – 43 = 48 – 43 = 5

now - a = mp = 5(24-a) or 4a = 120 or a= 30 hence a = 30 is largest

Note that you can make subscript 1 and 2 to distinguish the function of $f$ in latex as shown in the example below:

f_1 | $f_1$ |