Question

the roots of equation x² - (m-3) x + m = 0 are such that exactly one of them lies in the interval (1.2) then
1. 5<m<7
2. m<10
3. 2<m<5
4. m>10​

Answer 1

$\large\underline{\sf{Solution-}}$

$\large\underline{\sf{Given- }}$

A quadratic equation

$\red{\rm :\longmapsto\: {x}^{2} - (m - 3)x + m = 0}$

Let assume that

$\red{\rm :\longmapsto\:f(x) = {x}^{2} - (m - 3)x + m}$

Consider,

$\red{\rm :\longmapsto\:f(1) = {1}^{2} - (m - 3)(1) + m}$

$\rm \: = \: \: 1 - m + 3 + m$

$\rm \: = \: \: 4$

$\bf\implies \:f(1) = 4$

Consider,

$\red{\rm :\longmapsto\:f(2) = {(2)}^{2} - (m - 3)(2) + m}$

$\rm \: = \: \: 4 - 2m + 6 + m$

$\rm \: = \: \: 10 - m$

$\bf\implies \:f(2) = 10 - m$

Now, According to statement, it is given that one root exactly lies in between ( 1, 2 ).

$\rm :\implies\:f(1) \: f(2) \: < \: 0$

$\rm :\longmapsto\:4(10 - m) < 0$

$\rm :\longmapsto\:10 - m < 0$

$\rm :\longmapsto\: - m < - 10$

$\bf\implies \:m > 10$

Hence,

Option 4 is correct.

Additional Information :-

Let assume that

$\rm :\longmapsto\: \alpha , \beta \: are \: the \: roots \: of \: {ax}^{2} + bx + c = 0, \: then$

Case :- 1 Both roots are positive.

$\red{\rm :\longmapsto\: \alpha + \beta > 0}$

$\red{\rm :\longmapsto\: \alpha \beta > 0}$

$\red{\rm :\longmapsto\: {b}^{2} - 4ac \geqslant 0}$

Case :- 2 Both roots are negative

$\red{\rm :\longmapsto\: \alpha + \beta < 0}$

$\red{\rm :\longmapsto\: \alpha \beta > 0}$

$\red{\rm :\longmapsto\: {b}^{2} - 4ac \geqslant 0}$

Case :- 3 Roots are of opposite sign

$\red{\rm :\longmapsto\: \alpha \beta < 0}$

$\red{\rm :\longmapsto\: {b}^{2} - 4ac > 0}$