First slide

Theory of equations

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, then the absolute difference of the cubes of its roots is

Moderate

A

$10 \sqrt{5}$
B

$8 \sqrt{5}$
C

$8 \sqrt{3}$
D

$4 \sqrt{3}$

Solution

Given quadratic equation is
$(m^{2} + 1) x^{2} - 3 x + {(m^{2} + 1)}^{2} = 0 . . . (i)$
Let the roots of quadratic Eq. (i) are $α$ , and $β$ , so
$α + β = \frac{3}{m^{2} + 1} and αβ = m^{2} + 1$
According to the question, the sum of roots is greatest and it is possible only when "(m² + 1) is min" and "min value of m² + 1=1, when m =0".
$∴ α + β = 3 and αβ = 1, as m = 0$
Now, the absolute difference of the cubes of root
$\begin{array}{l} = |α^{3} - β^{3}| = | α - β | |α^{2} + β^{2} + αβ| \\ = \sqrt{(α + β)^{2} - 4 αβ} |(α + β)^{2} - αβ| \\ = \sqrt{9 - 4} | 9 - 1 | = 8 \sqrt{5} \end{array}$

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