Question

given that alpha and beta are the root of the equation 2 x square - 3 x + 4 is equal to zero find an equation whose roots are alpha + 1 upon alpha and beta + one upon beta​

Answer 1

Answer:

Step-by-step explanation:

2x^2+3x+4=0 …….(1)

And a,b are the roots of quadratic equation (1)

So, a+b= -3/2 and a×b = 4/2 = 2 …….(2)

Let a^3/b^3 and b^3/a^3 be the root of the required quadratic equation px^2+qx+r=0

So, sum of roots = -q/p

=> a^3/b^3 + b^3/a^3 = -q/p

=> (a^6 + b^6)/(a×b)^3 = -q/p

=> [(a^2 + b^2)(a^4+b^4-(a×b)^2)]/(a×b)^3 = -q/p

=>[{(a+b)^2-2ab}{((a+b)^2-2ab)^2–3(ab)^2}]/ (ab)^3 = -q/p

Substituting the value of equation (2) in above equation you will get

=>[{(-3/2)^2-2×2}{((-3/2)^2-2×2)^2–3×2^2}]/2^3 = -q/p

On soving you will get

=>[(-7/4)×(-143/16)]/8 = -q/p

=>q/p = -1oo1/512 …….(3)

And product of root = r/p

=>( a/b)^3 × (b/a)^3 = r/p

=>r/p = 1 …….(4)

The required quadratic equation will be

px^2+qx+r=0

=>x^2+(q/p)x+(r/p)=0

Put equation 3 and 4 in equation 5 you will get

=>x^2+(-1001/512)x+1=0

512x^2–1001x+512=0 ans