If m is chosen in the quadratic equation (m2 + 1)x2 - 3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to a √5, then a is

Question

If m is chosen in the quadratic equation (m2 + 1)x2 - 3x + (m2 + 1)2 = 0 such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to a √5, then a is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Sum of roots = (3/m2 + 1) ∵ sum of roots is greatest. ∴ m = 0 Hence equation becomes x2 - 3x + 1 = 0 Now, α + β = 3, α β = 1 ⇒ | - α - β| = √5 |α3 - β3| = |(α - β)(α2 + β2 + α β)| = √5 (9 - 1) = 8√5

Q. If $m$ is chosen in the quadratic equation
$(m^{2} + 1) x^{2} - 3 x + (m^{2} + 1)^{2} = 0$
such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to $a 5$ , then a is

Sum of roots $= \frac{3}{m ^{2} + 1}$
$∵$ sum of roots is greatest. $∴ m = 0$
Hence equation becomes $x^{2} - 3 x + 1 = 0$
Now, $α + β = 3, α β = 1$
$\Rightarrow ∣ - α - β ∣ = 5$
$∣ α^{3} - β^{3} ∣ = ∣ (α - β) (α^{2} + β^{2} + α β) ∣$
$= 5 (9 - 1) = 85$

Q. If m is chosen in the quadratic equation (m2+1)x2−3x+(m2+1)2=0 such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to a5​, then a is

Sum of roots =m2+13​ ∵ sum of roots is greatest. ∴m=0 Hence equation becomes x2−3x+1=0 Now, α+β=3,αβ=1 ⇒∣−α−β∣=5​ ∣α3−β3∣=∣(α−β)(α2+β2+αβ)∣ =5​(9−1)=85​

Q. If $m$ is chosen in the quadratic equation
$(m^{2} + 1) x^{2} - 3 x + (m^{2} + 1)^{2} = 0$
such that the sum of its roots is greatest and the absolute difference of the cubes of its roots is equal to $a 5$ , then a is

Sum of roots $= \frac{3}{m ^{2} + 1}$
$∵$ sum of roots is greatest. $∴ m = 0$
Hence equation becomes $x^{2} - 3 x + 1 = 0$
Now, $α + β = 3, α β = 1$
$\Rightarrow ∣ - α - β ∣ = 5$
$∣ α^{3} - β^{3} ∣ = ∣ (α - β) (α^{2} + β^{2} + α β) ∣$
$= 5 (9 - 1) = 85$