Tanishq's roots

By Vieta's Formulas, we know that the roots of the polynomial $f(a) = a^4-12a^3+17a^2+114a-216$ are $w,x,y,z$ . We note that $f(2) = 0$ and use this fact to help completely factor the polynomial into $f(a) = (a+3)(a-2)(a-4)(a-9)$ . Because $w\leq x \leq y \leq z$ , we have $(w,x,y,z) = (-3,2,4,9)$ .

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Now we consider $g(u)$ . Again, by Vieta's Formulas, we have $p+q+r = \frac{x}{w}, pq+qr+rp=\frac{y}{w}$ . Thus, $p^2+q^2+r^2 = (p+q+r)^2 - 2(pq+qr+rp) =$ $(\frac{x}{w})^2-2\frac{y}{w} = (\frac{2}{-3})^2 - 2\frac{4}{-3} = \frac{28}{9}$ , and so our answer is $28+9=\fbox{37}$ .

The beginning of this solution is the derivation of Vieta's formulas. Let $w, x, y, z$ be the roots of a monic 4th degree polynomial $f$ . Then $f(n) = (n-w)(n-x)(n-y)(n-z)\\=n^4-(w+x+y+z)n^3+(wx+wy+wz+xy+xz+yz)n^2-(wxy+wxz+wyz+xyz)n+wxyz\\=n^4-12n^3+17n^2+114n-216$ We now derive similar relationships for $g$ . $g(u) = w(u-p)(u-q)(u-r)=w(u^3-(p+q+r)u^2+(p+q+r)u-pqr)\\=wu^3+xu^2+yu+z$

Now we plug things into Vieta's formulas

We factor $f$ using standard techniques (guess & check worked well for me) $f(n)=(n+3)(n-2)(n-4)(n-9)\implies w=-3, x=2, y=4, z=9$

Finally, we solve the problem.

$p^2+q^2+r^2=(p+q+r)^2-2(pq+pr+qr)\\=\left(-\frac{x}{w}\right)^2-2\cdot\left(\frac{y}{w}\right)\\=\frac49+\frac83=\frac{28}9$ So $a+b=28+9=\boxed{37}$

According to Vietta's formulae, w,x,y,z is the roots of the equation $t^4-12t^3+17t^2+114t-216=0$ .

Solve for t, we obtain 4 roots: -3,2,4,9.

Therefore, $g(x)=-3u^3+2u^2+4u+9$ .

According to Vietta's formulae, $p+q+r=\frac{2}{3}$ and $pq+qr+rp=\frac{-4}{3}$ .

Therefore, $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+rp)=\frac{28}{9}$ .

The answer is: 28+9=37.

Clearly, the given system of equations can lead us to a polynomial of degree $4$ by Vieta's formulas. Using that concept, we get to:

$a^{4} - 12a^{3} + 17a^{2} + 114a - 216 = 0$

Where w, x, y and z are the roots of the polynomial.

At this point, by synthetic division we find the roots $-3, 2, 4, 9$ .

Now, our polynomial $g(u)$ has the form:

$g(u) = -3u^{3} + 2u^{2} + 4u + 9$

By Vieta's formulas (again), we have:

$p + q + r = \frac{2}{3}$ $pq + pr + rq = \frac{-4}{3}$

In the end, we just need a little bit of manipulation:

$p^{2} + q^{2} + r^{2} = (p+q+r)^{2} - 2(pq + pr + rq)$ $p^{2} + q^{2} + r^{2} = (\frac{2}{3})^{2} -2 \times \frac{-4}{3}$ $p^{2} + q^{2} + r^{2} = \frac{28}{9}$

Thus, our answer is $28 + 9 = 37$

From the given data and by using Vieta's formula, there exists a polynomial $f(a)$ such that

$f(a) = 0$ for $a = w,x,y,z$

Hence,

$f(a) = a^{4} - 12 a^{3} + 17a^{2}+114a - 216 = 0$

On solving you will get,

$f(a) = (a + 3)(a - 2)(a - 4)(a - 9) = 0$

As $w≤x≤y≤z$

$w = -3$

$x = 2$

$y = 4$

$z = 9$

Hence, $g(u)$ becomes as

$g(u) = -3u^{3} + 2u^{2} + 4u + 9$

by using Vieta's formula

$p + q + r = \frac{2}{3}$

$pq + qr + pr = -\frac{4}{3}$

Hence,

$p^{2} + q^{2} + r^{2} = (p + q + r)^{2} - 2(pq + qr + pr)$

$= ( \frac{2}{3})^{2} - 2(-\frac{4}{3})$

$= \frac{28}{9}$

Hence, $a + b = 28 + 9 = 37$

By Vietta's formulae,

w,x,y,z is the roots of the equation t^4−12t^3+17t^2+114t−216=0.

By SYNTHETIC DIVISION we found the roots -3,2,4,9.

Now g(x)=−3u^3+2u^2+4u+9.

Again using Vietta's Formoula p+q+r=2/3

and pq+qr+rp=−4/3.

=> p^2+q^2+r^2=(p+q+r)^2−2(pq+qr+rp)=(2/3)^2-2[(-4/3)]=28/9.

The answer is: 28+9=37

by Vietta's formulae, w,x,y,z is the roots of the equation

$t^{4}$ − $12t^{3}$ + $17t^{2}$ +114t−216=0.

by SYNTHETIC DIVISION we found the roots

-3,2,4,9.

now g(x)=−3 $u^{3}$ +2 $u^{2}$ +4u+9.

Again using Vietta's Formoula p+q+r=2/3 and pq+qr+rp=−4/3.

=> $p^{2}$ + $q^{2}$ + $r^{2}$ = $(p+q+r)^{2}$ −2(pq+qr+rp)= $(2/3)^{2}$ -2[(-4/3)]=28/9.

The answer is: 28+9=37.

Using Vieta's Formula, w,x,y,z are the roots of the equation. So simply by synthetic division, the roots we are are: -3, 2, 4, 9 g(x)=−3u^3+2u^2+4u+9 Using Vieta's Formula, p+q+r = 2/3 p^2+q^2+r^2 = (p+q+r)^2 − 2(pq+qr+rp) = (2/3)^2 - 2(-4/3) = 28/9. a + b = 28 + 9 = 37