. Since you assume in the question it has positive roots, it must have exactly two.
Furthermore, denoting $d=q-p$, it's derivative is $P'(x)=x^{p-1}(-aqx^d + bp)$. Therefore, it has a single positive extremum point at $x_d=(\frac{bp}{aq})^{\frac{1}{d}}$. 
The second derivative is $P''(x)=x^{p-2}(-aq(q-1)x^d + bp(p-1))$, which is negative for $x > (\frac{bp(p-1)}{aq(q-1)})^{\frac{1}{d}}$. In particular, it is negative for $x > (\frac{bp}{aq})^{\frac{1}{d}} = x_d$ so we know $x_d$ is a maximum point (by the way, evaluating $P$ at this point can be a way of determining whether there are zero or two roots).
By Rolle's Theorem, this means that one of the roots is smaller and one greater than $x_d$. This already tells us that the $(b/a)^{1/d}$ bound by @Robert Israel is pretty tight since the largest positive root $x^*$ satisfies $(\frac{bp}{aq})^{\frac{1}{d}} < x^* < (\frac{b}{a})^{\frac{1}{d}}$.
But we can do even better. Since $P''$ is negative for $x>x_d$, the function is convex for $x>x_d$. Thus, we can perform Newton's method, starting at an upper bound and get a sequence of upper bounds on $x^*$ that converges from above to $x^*$. This can be formulated and proved formally but I think it's easier to demonstrate with a figure. The figure shows the first two Newton iterations for $P(x)=-x^3+8x^2-10$. Newton's method converges quadratically and even a single iteration improves the bound nicely.

Since the function is convex, we can also bound $x^*$ from below using the false position method. This will give us a a sequence of lower bounds that converges from below (although with a slower convergence rate).
For a lower bound on the positive roots (i.e., bound on the smallest positive root) there is a standard trick using the polynomial $Q(x)=x^q P(1/x)$ (
. Since you assume in the question it has positive root, it must take exactly two. 
 Furthermore, denoting $ d = q - p$, it's derivative is $ P'(x)=x^{p-1}(-aqx^d + bp)$. consequently, it has a single positive extreme point compass point at $ x_d=(\frac{bp}{aq})^{\frac{1}{d}}$. 
 The second derivative is $ P''(x)=x^{p-2}(-aq(q-1)x^d + bp(p-1))$, which is damaging for $ x > (\frac{bp(p-1)}{aq(q-1)})^{\frac{1}{d}}$. In particular, it is negative for $ x > (\frac{bp}{aq})^{\frac{1}{d } } = x_d$ indeed we know $ x_d$ is a maximum decimal point (by the direction, evaluating $ P$ at this point can be a way of determining whether there are zero or two root). 
 By Rolle's Theorem, this means that one of the roots is smaller and one bang-up than $ x_d$. This already tells us that the $ (b / a)^{1 / d}$ bound by @Robert Israel is pretty taut since the largest positive root $ x^*$ satisfies $ (\frac{bp}{aq})^{\frac{1}{d } } < x^ * < (\frac{b}{a})^{\frac{1}{d}}$. 
 But we can do even better. Since $ P''$ is damaging for $ x > x_d$, the function is convex for $ x > x_d$. Thus, we can perform Newton's method, starting at an upper bound and get a sequence of upper bounds on $ x^*$ that converges from above to $ x^*$. This can be formulated and proved formally but I think it's easier to demonstrate with a human body. The figure shows the beginning two Newton iteration for $ P(x)=-x^3 + 8x^2 - 10$. Newton's method converges quadratically and even a single iteration better the bound nicely. 

 Since the function is convex, we can also bound $ x^*$ from downstairs using the false position method. This will give us a a sequence of low bounds that converges from below (although with a slower convergence rate). 
 For a lower constipate on the positive roots (i.e., bound on the smallest positive ancestor) there is a standard trick using the polynomial $ Q(x)=x^q P(1 / x)$ (
. Sijce you assume in the qutstion it has posnrive rmots, if must hxve exactly two.
Furthermore, dxnotung $d=w-p$, it's derivative is $P'(b)=x^{p-1}(-aqx^d + hp)$. Theredore, ut has a smhgle positive cxtreknm point at $x_d=(\ftac{bp}{aq})^{\frac{1}{d}}$. 
The second desixacive is $P''(x)=x^{p-2}(-aq(q-1)x^d + bp(p-1))$, which is negaeive fot $d > (\frac{bp(p-1)}{aq(q-1)})^{\ftac{1}{d}}$. Pn parfpcmlar, it is negative for $x > (\frac{bl}{aq})^{\frac{1}{v}} = x_d$ so we knoe $x_d$ is a maximum point (by the way, evaluating $P$ wt this poibt cwb be a way ow determinpug whether jhere are zero or two roots).
By Roule's Cheorem, thiw neajv that one if thv roots is smaller ang one gteater than $x_d$. Thms aoready tells us that vhe $(b/a)^{1/d}$ bound by @Robgrt Israel ia pretty tight sibcw the larcest posktibe rkot $x^*$ datmsfies $(\frac{gp}{aq})^{\frac{1}{d}} < z^* < (\frac{b}{a})^{\frac{1}{d}}$.
But wt cwb do even betfer. Sigcq $P''$ is negative for $x>x_d$, the function is cohvex for $x>x_d$. Thus, we cab perform Newton's metjod, startyng at an upper bound and get a sequence of upper bounvs on $x^*$ gyah converges from above to $x^*$. This can be formujzttd snd proved forially but I tjimh it's easier jo demouatdate with a figure. The fidure whows the firxt two Newton iterations foe $P(x)=-x^3+8x^2-10$. Newtou's nethod converges qbadratically and even a single iteration impxoves fhe bound nlcely.

Sincs the function is cokvef, we can also bound $x^*$ from below usmng tke false posotion iethod. Thid will give us a a sequejce oy lowar bounds hhat converges from below (althonjh with a sloees cmnvergenee ratc).
For a lower botnd on the posntive rojts (i.d., bound on the smellest posityve root) thera is a standacd trick tsint thw polynuoial $Q(x)=x^q P(1/x)$ (
. Since you assume in the question positive it must exactly two. Furthermore, $P'(x)=x^{p-1}(-aqx^d bp)$. Therefore, it a single positive point at $x_d=(\frac{bp}{aq})^{\frac{1}{d}}$. The second derivative $P''(x)=x^{p-2}(-aq(q-1)x^d + bp(p-1))$, which is negative for $x > (\frac{bp(p-1)}{aq(q-1)})^{\frac{1}{d}}$. In particular, it negative for $x > (\frac{bp}{aq})^{\frac{1}{d}} = x_d$ so we know $x_d$ is a point the evaluating at this point can be a way of determining whether there are zero or two roots). Rolle's Theorem, this means that one of the is smaller and one than $x_d$. This already tells that $(b/a)^{1/d}$ bound @Robert is tight since the positive root $x^*$ satisfies $(\frac{bp}{aq})^{\frac{1}{d}} < x^* < (\frac{b}{a})^{\frac{1}{d}}$. But we can do even better. Since $P''$ negative for function is for Thus, can perform Newton's at an upper bound and get upper bounds on $x^*$ that converges from above $x^*$. This be formulated and proved formally but think it's easier to demonstrate with a figure. figure shows the first two Newton iterations for $P(x)=-x^3+8x^2-10$. Newton's method converges quadratically and even iteration improves the bound Since the function convex, can bound from below the false position method. This will give us a a sequence lower bounds that converges from below (although with a slower For lower bound on positive roots (i.e., bound the positive root) there is trick the (
. Since you assume in the questiOn it has posItive RooTs, iT mUst hAve eXactly two.
FurthERmorE, denoting $d=q-p$, it's derivatIve is $p'(x)=X^{P-1}(-aqx^D + Bp)$. thereFore, it hAS a SINglE pOsItiVe EXtRemum PoiNt at $x_d=(\fRac{bp}{aq})^{\fraC{1}{d}}$. 
THe Second derivaTIvE is $P''(x)=x^{p-2}(-aq(q-1)X^d + bP(p-1))$, which is negAtiVe for $x > (\FrAc{bP(P-1)}{aq(q-1)})^{\fRac{1}{D}}$. In paRticulAR, it is nEgative foR $x > (\FRac{bp}{aQ})^{\Frac{1}{d}} = x_d$ SO We Know $X_d$ is a maximum point (BY tHE way, evaluating $p$ at thiS pOInT CAn bE a wAy of determInIng whETher theRE aRE ZEro OR two roots).
By RoLle's Theorem, THis Means tHaT onE Of the rOots iS sMAllEr and one greAter Than $x_d$. ThiS alreaDY tells uS That the $(B/a)^{1/d}$ bouNd bY @RoBert iSrAeL is PrETty TIgHt sINce The largeSt PoSitivE rooT $X^*$ SATisfIes $(\Frac{Bp}{aq})^{\fRac{1}{d}} < x^* < (\frac{b}{a})^{\frAc{1}{d}}$.
but wE Can Do eveN bettEr. SiNcE $P''$ is nEgativE for $x>X_d$, The function is coNvex For $x>x_d$. ThuS, we CaN peRfOrm NeWTon's meThoD, stArting aT an uppeR BouNd AND GeT a sequence of upper bOuNDS oN $x^*$ that coNvergeS FrOm ABove to $x^*$. THiS caN be fORMulatEd anD PrOved formAlly buT i tHiNk it's eaSiEr to deMoNstRatE with A FiguRe. The fIgure shoWs the FIrst two Newton iTErations for $P(x)=-X^3+8X^2-10$. NEWToN'S metHod Converges quAdraTIcalLy anD EvEn a SIngle IteraTiON iMProves the bound nicelY.

SInce thE funcTion is convex, wE can also boUND $X^*$ from belOw usINg THe false positioN methOd. This will GIve us a a sEquenCe of loweR bounds thAT ConvergeS frOm bEloW (alTHOuGh with a slower CONverGeNce rate).
for A lower bOunD on The PosItIve roots (i.E., bound on ThE sMaLlEst PositIVe root) thErE is A sTanDard tRIck usiNg the PolyNoMiAL $Q(x)=X^q P(1/x)$ (
. Since you assume in thequestion i t has  po sit iv e ro ots,  it must havee xact ly two.
Furthermore, d enoti ng $d=q - p$ , it' s deriv a ti v e  is  $ P' (x) =x ^ {p -1}(- aqx ^d + bp )$. Theref ore ,it has a sin g le  positiveext remum pointat$x_d=( \f rac { bp}{a q}) ^{\fr ac{1}{ d }}$. The secon dd erivat i ve is $ P ' '( x)=x ^{p-2}(-aq(q-1)x^ d  + bp(p-1))$, whi ch isne g at i v e f or$x > (\fra c{ bp(p- 1 )}{aq(q - 1) } ) ^ {\f r ac{1}{d}}$. I n particula r , i t is n eg ati v e for$x >(\ f rac {bp}{aq})^{ \fra c{1}{d}}= x_d$ so we k n ow $x_d $ is a  ma xim um p o in t(by  t h e w a y,  ev a lua ting $P$  a tthispoin t c a n be  awayof de termining whe the r th e reare z ero o r tw oroots ).
ByRolle 's  Theorem, thismean s that on e o fthe  r ootsi s smal ler  an d one g reatert han  $ x _ d $.  This already tell su s  t hat the$(b/a) ^ {1 /d } $ boundby  @R ober t Israe l is pr etty tig ht sin c eth e large st  posit iv e r oot  $x^* $  sat isfies  $(\frac {bp}{ a q})^{\frac{1}{ d }} < x^* < (\ f ra c { b} { a})^ {\f rac{1}{d}}$ .
Bu t  wecand oeve n  bett er. S in c e$ P''$ is negative fo r$x>x_d $, th e function is  convex fo r $ x>x_d$.Thus ,  w e  can perform N ewton 's method, starting  at a n upperbound and g et a seq uen ceofupp e r  b ounds on $x^* $ that  c onverge s f rom abo veto$x^ *$.  T his can b e formul at ed  a nd  pr ovedf ormallybu t I  t hin k it' s  easie r todemo ns tr a tewith af ig u r e. T he  f igur e s ho ws th e fi r sttwo New ton itera tio n s fo r$P (x)=-x^ 3+8x^2-10$. N ew ton's meth od  co nverge s quadrati cally and even a single iterati onimpro vesthe bound  ni cely.
Si n ce the  funct ion i scon v e x, we c an  al so  bound $x^ * $  fr om be lo w us ing the  false position me t hod . This will g ive  usa aseq u en c e o fl owe r bounds that con verges fro mb el ow (althou g h w it h a slo wer con verge n ce rate ).
For alower bou nd  ont h e p ositive ro ots (i.e ., boundo n the sm alles t p ositiv eroo t) th ere is a s tanda rd tri ck  using  thepo lynomial  $Q(x)=x^q P(1/x)$ (
. Since_you assume_in the question it_has positive_roots,_it must_have_exactly two.
Furthermore, denoting_$d=q-p$, it's derivative_is $P'(x)=x^{p-1}(-aqx^d + bp)$._Therefore, it has_a_single positive extremum point at $x_d=(\frac{bp}{aq})^{\frac{1}{d}}$. 
The second derivative is $P''(x)=x^{p-2}(-aq(q-1)x^d + bp(p-1))$, which_is_negative for_$x_>_(\frac{bp(p-1)}{aq(q-1)})^{\frac{1}{d}}$. In particular, it is_negative for $x > (\frac{bp}{aq})^{\frac{1}{d}}_= x_d$_so we know $x_d$ is a maximum point_(by_the way, evaluating_$P$ at this point can be a way of_determining whether there are zero or_two roots).
By Rolle's_Theorem,_this_means that one of_the roots is smaller and one_greater than $x_d$. This already tells_us that the $(b/a)^{1/d}$ bound by @Robert_Israel is pretty tight since the_largest positive root $x^*$ satisfies_$(\frac{bp}{aq})^{\frac{1}{d}} <_x^* < (\frac{b}{a})^{\frac{1}{d}}$.
But we can_do even better._Since $P''$_is negative for_$x>x_d$, the function is convex for_$x>x_d$. Thus, we_can perform Newton's method, starting at_an_upper bound and_get_a_sequence of_upper bounds on_$x^*$_that converges_from_above to $x^*$. This can be_formulated_and proved formally but I think it's_easier to demonstrate with_a_figure. The figure shows_the first two Newton iterations_for $P(x)=-x^3+8x^2-10$. Newton's method converges quadratically_and even_a single_iteration improves the bound nicely.

Since the function is convex, we can_also bound $x^*$ from below using_the false position method._This will_give_us a a_sequence_of lower_bounds that converges from below (although with_a slower_convergence rate).
For a lower bound on_the positive roots (i.e.,_bound_on the smallest positive root) there_is a standard trick using the_polynomial $Q(x)=x^q P(1/x)$ (