Question

find the condition for m for which the roots of a quadratic equation 4x square - 5 M X plus one equal to zero are real

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,

Given:

$4 {x}^{2} - 5Mx + 1 = 0$

To find $\textbf{M}$

$4 {x}^{2} - 5Mx + 1 = 0$

Subtracting " 4x square " or " 4x^2 " from left and right hand sides,,,,

$4 {x}^{2} - 5Mx + 1 - 4 {x}^{2} = 0 - 4 {x}^{2}$

Simplifying the given equation, ,, we get,,,,

$- 5Mx + 1 = - 4 {x}^{2}$

Subtracting "1" from left and right hand sides,,,,,

$- 5Mx + 1 - 1 = - 4 {x}^{2} - 1$

Again Simplifying the given equation,,,,,,

$- 5Mx = - 4 {x}^{2} - 1$

Dividing both the sides by "-5x" with the condition of "x is not equal to zero"

$\frac{ - 5Mx}{ - 5x} = \frac{4 {x}^{2} }{ - 5x} - \frac{1}{ - 5 {x}^{2} } \\ \\ x \: is \: not \: equal \: to \: zero \:$

Simplifying the whole equation given,,

Therefore,,,,,,

$= > \frac{5xM}{5x} = \frac{ - 4 {x}^{2} - 1}{ - 5x} \\ \\ \\ = > \frac{xM}{x} = - \frac{ - 4 {x}^{2} - 1 }{5x} \\ \\ \\ = > M = - \frac{ - 4 {x}^{2} - 1 }{5x} \\ \\ x \: is \: not \: equal \: to \: zero$

HOPE THIS HELPS YOU AND CLEARS THE DOUBT FOR "SOLVE FOR M" IN QUADRATIC EQUATIONS!!!!!!