One root of the quadratic equation `x^(2)-12x+a=0` is thrice the other. Find the value of a?

Answer

`a epsilon (3/4,oo)`

Explanation

Given equation is
`x^(4)+2ax^(3)+x^(2)+2ax+1=0`…i
On dividing by `x^(2)` we get
`x^(2)+2ax+1+(2a)/x+1/(x^(2))=0`
`implies(x^(2)+1/(x^(2))+2a(x+1/x)+1=0`
`implies(x+1/x)^(2)-2+2a(x+1/x)+1=0`
or `(x+1/x)^(2)+2a(x+1/x)-1=0`
or `y^(2)-2ay-1=0`, where `y=x+1/x`
`:.y=(-2a+-sqrt((4a^(2)+4)))/2=-a+-sqrt((a^(2)+1))`
Taking `+` sign we get
`y=-a+sqrt((a^(2)+1))`
`impliesx+1/x=-a+sqrt((a^(2)+1))`
or `x^(2)+(a-sqrt((a^(2)+1)))x+1=0`..........ii
Taking `-` sign we get `y=-a-sqrt((a^(2)+1))`
`impliesx+1/x=-a-sqrt((a^(2)-1))`
or `x^(2)+(a+sqrt((a^(2)+1))x+1=0`...........iii
Let `alpha, beta` be the roots of Eq. (ii) and `gamma, delta` be the roots of Eq. (iii)
Then `alpha+beta=sqrt((a^(2)+1))-a`
and `alpha beta=1`
and `gamma+delta=-sqrt((a^(2)+1))-a`
and `gamma delta=1`
Clearly `alpha +betagt0` and `alpha betagt0`
`:.` Either `alpha, beta` will be imaginary or both real and positive according to the Eq. (i) has atleast two distinct negative roots.
Therefore both `gamma` and `delta` must be negative. Therefore
(i) `gamma delta gt0` which is true as `gamma delta=1`
`implies(-a+sqrt((a^(2)+1)))lt0`
`impliesa+sqrt((a^(2)+1))gt0` which is true for all a.
`:. a epsilonR`
(iii) `Dgt0`
`:.(a+sqrt((a^(2)+1)))^(2)-4gt0`
`implies(a+sqrt((a^(2)+1)+2)(a+sqrt((a^(2)+1))-2)gt0`
`:.a+sqrt((a^(2)+1))+2gt0`
`:.a+sqrt((a^(2)+1))-2gt0`
`impliessqrt((a^(2)+1))gt2-a`
`{(age2),("or"a^(2)+gt(2-a)^(2),"if"alt02:}`
`implies{*age2),("or"agt3/4,"if"alt2):}`
`implies{(age2),("or"3/4ltalt2):}`
Hence `3/4ltaltoo` or `a epsilon(3/4,oo)`