Question

If one root of the equation $(1 - m){x^2} + 1x + 1 = 0$ is double of the other and is real, find the greatest value of m?

Answer 1

The given question is of quadratic equation in which it is given that the roots are real, here we need to use sridharacharya rule in order to write the equation for the roots, and then according to the conditions we need to get the maximum value for the “m” as per question.
Formulae Used: For the roots of the quadratic equation, of general equation say:
$\Rightarrow a{x^2} + bx + c = 0$
Determinant “d” can be written as:
$\Rightarrow d = \sqrt {{b^2} - 4ac}$
Roots of the equations are:
$\Rightarrow roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step-by-step solution:
Here in the given question we first need to solve for the roots and then as per the instruction in the question we need to solve for the values, on solving we get:

\Rightarrow roots = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ \Rightarrow roots = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times (1 - m)c} }}{{2 \times (1 - m)}} $$ In the question we need to understand that the roots are real, hence the value under the square root will have to be positive, and since denominator cannot become zero hence value of “m” will be greater than one, on solving we get: Condition 1: $$ \Rightarrow \sqrt {{1^2} - 4 \times (1 - m) \times 1} > 0 \\\ \Rightarrow (1 - m) \leqslant 0 \\\ \Rightarrow m \geqslant 1 $$ Now, Condition 2: $$ \Rightarrow 2 \times (1 - m) \ne 0 \\\ \Rightarrow m \ne 1 $$ **From condition 1 and 2 we can say the value of “m” can be anyone's value greater than one.** **Note:** To solve questions in which a range of values, need to be determined then we need to put some equations and then solve accordingly to get the optimum value of range for the equation or variable, here also we first make equation and then get the range of values for the asked variable.