SATHEE: Theory of Equations 1 Question 3

Theory of Equations 1 Question 3

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

(a) $10 \sqrt{5}$

(b) $8 \sqrt{5}$

(d) $4 \sqrt{3}$

(2019 Main, 9 April II)

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Answer:

Correct Answer: 3. (b)

Solution:

Given quadratic equation is

$(m^{2} + 1) x^{2} - 3 x + {(m^{2} + 1)}^{2} = 0$

Let the roots of quadratic Eq. (i) are $α$ and $β$ , so $α + β = \frac{3}{m^{2} + 1}$ and $α β = m^{2} + 1$

According to the question, the sum of roots is greatest and it is possible only when " $(m^{2} + 1)$ is minimum" and “minimum value of $m^{2} + 1 = 1$ , when $m = 0$ “.

$∴ α + β = 3$ and $α β = 1$ , as $m = 0$

Now, the absolute difference of the cubes of roots

$\begin{aligned} = | α^{3} - β^{3} | \\ = | α - β | | α^{2} + β^{2} + α β | \\ = \sqrt{(α + β)^{2} - 4 α β} | (α + β)^{2} - α β | \\ = \sqrt{9 - 4} | 9 - 1 | = 8 \sqrt{5} \end{aligned}$

Theory of Equations 1 Question 3

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

Answer:

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Theory of Equations 1 Question 3

3. 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

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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