Question

Please solve it with full explanation

Answer 1

Refer to attachment !!!!!:-)

Answer 2

HELLO FRIEND HERE IS YOUR ANSWER,,,,,

Given equation,,

$m {x}^{2} - 6mx + 5m + 1 > 0$

If the values of variable "x" are real then,,, by applying the calculation of discriminant,, that is,,

${b}^{2} - 4ac > 0$

of the given quadratic equation, it becomes,,

$m - 6m + (5m + 1)$

Here,,,

$b = 6m \: \: \: \: \: \: \: a = m \: \: \: \: \: \: c = 5m + 1$

Therefore by substituting the above values we get,,,

${(6m)}^{2} - 4(m) \times (5m + 1) > 0$

Remove parentheses,, that is,, (a) = a

${(6m)}^{2} - 4m(5m + 1) > 0$

Applying exponent rule, that is,

${(a \times b)}^{n} = {a}^{n} {b}^{n} \\ \\ \\ {6}^{2} {m}^{2} - 4m(5m + 1) > 0 \\ \\ \\ {36}^{2} - 4m(5m + 1) > 0$

Applying distributive law, that is,

$a(b + c) = ab + ac \: \: \: \: \: \: \: here \\ \\ \\ a = - 4m \: \: \: \: \: \: \: b = 5m \: \: \: \: \: c = 1$

${36m}^{2} - 4m \times 5m + ( - 4m) \times 1 > 0 \\ \\ \\ {36m}^{2} - 4 \times 5mm - 4 \times 1 \times m > 0 \\ \\ \\ {36m}^{2} - 20mm - 4m > 0 \\ \\ \\ {36m}^{2} - {20m}^{2} - 4m > 0 \\ \\ \\ {16m}^{2} - 4m > 0$

Dividing by the value of "4" on left hand side,,

${4m}^{2} - m > 0$

Isolating the given variable "m"

$m(4m - 1) > 0$

Therefore, If m > 0, the expression is positive then,,

$m < \frac{1}{4}$

Therefore, if m < 0, the expression may presume positive charges or negative charges for the values of real "x".

Thus the final answer for this particular query is,,,,

$0 \leqslant m < \frac{1}{4}$

HOPE IT HELPS AND CLEARS YOUR DOUBTS FOR INEQUALITIES BY QUADRATIC FORM!!!!