Let α and β are the roots of equation ax^2+bx+c=0 a≠0. If 1, α+β, αβ are in arithmetic progression and α, 2, β are in harmonic progression, then the…

Question

Let α and β are the roots of equation ax^2+bx+c=0 a≠0. If 1, α+β, αβ are in arithmetic progression and α, 2, β are in harmonic progression, then the value of (α^2+β^2-2α^2β^2)/(2α^2+β^2) is equal to

MARKS Content Team · Accepted Answer

Correct answer: (D) 1.5. 1, α+β,αβ are in A.P. ⇒1,(-b)/a,c/a are in A.P. ⇒1+c/a=(-2b)/a⇒a+c+2b=0 … 1 1/(α),1/2,1/(β) are in A.P. ⇒1/(α)+1/(β)=1⇒α+β=αβ ⇒(-b)/a=c/a⇒b+c=0 … 2 From 1 & 2 we get, a=-b=c ⇒α,β are roots of equation x^2-x+1=0 Now, (α^2+β^2-2α^2β^2)/(2α^2+β^2)=1/2-((αβ)^2)/((α+β)^2-2αβ) =1/2-(1^2)/(1^2-21)=1/2+1=1.5

Let $α$ and $β$ are the roots of equation $a x^{2} + b x + c = 0 (a \neq 0) .$ If $1, α + β, α β$ are in arithmetic progression and $α, 2, β$ are in harmonic progression, then the value of $\frac{α^{2} + β^{2} - 2 α^{2} β^{2}}{2 (α^{2} + β^{2})}$ is equal to

Solution

More Quadratic Equation Questions

Let α and β are the roots of equation ax2+bx+c=0 a≠0. If 1, α+β, αβ are in arithmetic progression and α, 2, β are in harmonic progression, then the value of α2+β2-2α2β22α2+β2 is equal to

Solution

More Quadratic Equation Questions

Let $α$ and $β$ are the roots of equation $a x^{2} + b x + c = 0 (a \neq 0) .$ If $1, α + β, α β$ are in arithmetic progression and $α, 2, β$ are in harmonic progression, then the value of $\frac{α^{2} + β^{2} - 2 α^{2} β^{2}}{2 (α^{2} + β^{2})}$ is equal to