Question

If a and b are the roots of x^2 -3x+p = 0 and c, d are roots of x^2 -12x+q = 0 , where a, b, c, d form a G.P. prove that (q + p) : (q –p) = 17 : 15

Answer 1

Given:

$\sf{\leadsto{The \ roots \ of \ x^{2}-3x+p=0 \ are}}$

$\sf{a \ and \ b.}$

$\sf{\leadsto{The \ roots \ of \ x^{2}-12x+q=0 \ are}}$

$\sf{c \ and \ d.}$

$\sf{\leadsto{a, \ b, \ c, \ d \ form \ a \ G.P.}}$

To prove:

$\sf{\longmapsto{(q+p):(q-p)=17:15}}$

Solution:

$\sf{If \ quadratic \ equation \ x^{2}-3x+p=0}$

$\sf{has \ a \ and \ b \ as \ it's \ roots. }$

$\sf{Then,}$

$\sf{\mapsto{a+b=3...(1)}}$

$\sf{\mapsto{ab=p...(2)}}$

$\sf{If \ quadratic \ equation \ x^{2}-12x+q=0}$

$\sf{has \ c \ and \ d \ as \ it's \ roots. }$

$\sf{Then,}$

$\sf{\mapsto{c+d=12...(3)}}$

$\sf{\mapsto{cd=q...(4)}}$

$\sf{But, \ a, \ b, \ c \ and \ d \ are \ in \ G.P.}$

$\boxed{\sf{a_{n}=ar^{n-1}}}$

$\sf{Hence,}$

$\sf{First \ term \ (a_{1})=a=a,}$

$\sf{Second \ term \ (a_{2})=b=ar,}$

$\sf{Third \ term \ (a_{3})=c=ar^{2},}$

$\sf{Forth \ term \ (a_{4})=d=ar^{3}}$

$\sf{In \ a \ G.P. \ \dfrac{a_{n+1}}{a_{n}} \ is \ constant}$

$\sf{\therefore{\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}}}$

$\sf{\therefore{\dfrac{b}{a}=\dfrac{d}{c}}}$

$\sf{By \ componendo}$

$\sf{\dfrac{a+b}{a}=\dfrac{c+d}{c}}$

$\sf{...from \ (1) \ and (3) \ we \ get}$

$\sf{\dfrac{3}{a}=\dfrac{12}{c}}$

$\sf{\therefore{\dfrac{c}{a}=\dfrac{12}{3}}}$

$\sf{But, \ a=a \ and \ c=ar^{2}}$

$\sf{\therefore{\dfrac{ar^{2}}{a}=\dfrac{12}{3}}}$

$\sf{\therefore{r^{2}=4}}$

$\sf{\therefore{r=\pm2}}$

____________________________________

$\sf{When \ r=2}$

$\sf{a=a,}$

$\sf{b=ar=2a,}$

$\sf{c=ar^{2}=4a,}$

$\sf{d=ar^{3}=8a}$

$\sf{Substituting \ the \ values \ of \ a \ and \ b \ in}$

$\sf{equation(1)}$

$\sf{a+2a=3}$

$\sf{\therefore{3a=3}}$

$\boxed{\sf{\therefore{a=1}}}$

$\sf{From \ equation(2)}$

$\sf{ab=p}$

$\sf{But, \ a=1 \ and \ r=2}$

$\sf{\therefore{p=2}}$

$\sf{From \ equation(4)}$

$\sf{cd=q}$

$\sf{\therefore{(ar^{2})(ar^{3})=q}}$

$\sf{But, \ a=1 \ and \ r=2}$

$\sf{\therefore{(1^{2})(2^{5})=q}}$

$\sf{\therefore{q=32}}$

$\sf{\leadsto{\dfrac{q+p}{q-p}=\dfrac{32+2}{32-2}}}$

$\sf{\therefore{\dfrac{q+p}{q-p}=\dfrac{34}{30}}}$

$\sf{\therefore{\dfrac{q+p}{q-p}=\dfrac{17}{15}}}$

$\sf{\longmapsto{\therefore{(q+p):(q-p)=17:15}}}$

____________________________________

$\sf{When \ r=-2}$

$\sf{a=a,}$

$\sf{b=ar=-2a,}$

$\sf{c=ar^{2}=4a,}$

$\sf{d=ar^{3}=-8a}$

$\sf{Substituting \ the \ values \ of \ a \ and \ b \ in}$

$\sf{equation(1)}$

$\sf{a-2a=3}$

$\sf{\therefore{-a=3}}$

$\boxed{\sf{\therefore{a=-3}}}$

$\sf{From \ equation(2)}$

$\sf{ab=p}$

$\sf{\therefore{a(ar)=p}}$

$\sf{\therefore{a^{2}r=p}}$

$\sf{But, \ a=-3 \ and \ r=-2}$

$\sf{\therefore{p=(-3)^{2}(-2)}}$

$\sf{\therefore{p=-18}}$

$\sf{From \ equation(4)}$

$\sf{cd=q}$

$\sf{\therefore{(ar^{2})(ar^{3})=q}}$

$\sf{\therefore{a^{2}r^{5}=q}}$

$\sf{But, \ a=-3 \ and \ r=-2}$

$\sf{\therefore{q=(-3)^{2}(-2)^{5}}}$

$\sf{\therefore{q=9(-32)}}$

$\sf{\therefore{q=-288}}$

$\sf{\leadsto{\dfrac{q+p}{q-p}=\dfrac{-288+(-18)}{-288-(-18)}}}$

$\sf{\therefore{\dfrac{q+p}{q-p}=\dfrac{-306}{-270}}}$

$\sf{\therefore{\dfrac{q+p}{q-p}=\dfrac{18(17)}{18(15)}}}$

$\sf{\therefore{\dfrac{q+p}{q-p}=\dfrac{17}{15}}}$

$\sf{\longmapsto{\therefore{(q+p):(q-p)=17:15}}}$

$\sf{Hence \ proved,}$

$\sf\purple{\tt{(q+p):(q-p)=17:15}}$

Answer 2

Answer:

The roots of x2−3x+p=0 are

\sf{a \ and \ b.}a and b.

\sf{\leadsto{The \ roots \ of \ x^{2}-12x+q=0 \ are}}⇝The roots of x2−12x+q=0 are

\sf{c \ and \ d.}c and d.

\sf{\leadsto{a, \ b, \ c, \ d \ form \ a \ G.P.}}⇝a, b, c, d form a G.P.

To prove:

\sf{\longmapsto{(q+p):(q-p)=17:15}}⟼(q+p):(q−p)=17:15

Solution:

\sf{If \ quadratic \ equation \ x^{2}-3x+p=0}If quadratic equation x2−3x+p=0

\sf{has \ a \ and \ b \ as \ it's \ roots. }has a and b as it′s roots.

\sf{Then,}Then,

\sf{\mapsto{a+b=3...(1)}}↦a+b=3...(1)

\sf{\mapsto{ab=p...(2)}}↦ab=p...(2)

\sf{If \ quadratic \ equation \ x^{2}-12x+q=0}If quadratic equation x2−12x+q=0

\sf{has \ c \ and \ d \ as \ it's \ roots. }has c and d as it′s roots.

\sf{Then,}Then,

\sf{\mapsto{c+d=12...(3)}}↦c+d=12...(3)

\sf{\mapsto{cd=q...(4)}}↦cd=q...(4)

\sf{But, \ a, \ b, \ c \ and \ d \ are \ in \ G.P.}But, a, b, c and d are in G.P.

\boxed{\sf{a_{n}=ar^{n-1}}}an=arn−1

\sf{Hence,}Hence,

\sf{First \ term \ (a_{1})=a=a,}First term (a1)=a=a,

\sf{Second \ term \ (a_{2})=b=ar,}Second term (a2)=b=ar,

\sf{Third \ term \ (a_{3})=c=ar^{2},}Third term (a3)=c=ar2,

\sf{Forth \ term \ (a_{4})=d=ar^{3}}Forth term (a4)=d=ar3

\sf{In \ a \ G.P. \ \dfrac{a_{n+1}}{a_{n}} \ is \ constant}In a G.P. anan+1 is constant

\sf{\therefore{\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}}}∴ab=bc=cd

\sf{\therefore{\dfrac{b}{a}=\dfrac{d}{c}}}∴ab=cd

\sf{By \ componendo}By componendo

\sf{\dfrac{a+b}{a}=\dfrac{c+d}{c}}aa+b=cc+d

\sf{...from \ (1) \ and (3) \ we \ get}...from (1) and(3) we get

\sf{\dfrac{3}{a}=\dfrac{12}{c}}a3=c12

\sf{\therefore{\dfrac{c}{a}=\dfrac{12}{3}}}∴ac=312

\sf{But, \ a=a \ and \ c=ar^{2}}But, a=a and c=ar2

\sf{\therefore{\dfrac{ar^{2}}{a}=\dfrac{12}{3}}}∴a