Question

if x²-5x+8 is positive, then find range of x​

Answer 1

SOLUTION

The range of x when

$\sf{ {x}^{2} - 5x + 8 \: \: \: is \: positive }$

EVALUATION

Here the given expression is

$\sf{ {x}^{2} - 5x + 8 \: }$

We now simplify it as below

$\sf{ {x}^{2} - 5x + 8 \: }$

$\displaystyle \sf{ = {x}^{2} -2 .x . \frac{5}{2} + { \bigg( \frac{5}{2} \bigg)}^{2} - { \bigg( \frac{5}{2} \bigg)}^{2} + 8 \: }$

$\displaystyle \sf{ = { \bigg(x - \frac{5}{2} \bigg)}^{2} + 8 - \frac{25}{4} \: }$

$\displaystyle \sf{ = { \bigg(x - \frac{5}{2} \bigg)}^{2} + \frac{32 - 25}{4} \: }$

$\displaystyle \sf{ = { \bigg(x - \frac{5}{2} \bigg)}^{2} + \frac{7}{4} \: }$

$\displaystyle \sf{ \because { \bigg(x - \frac{5}{2} \bigg)}^{2} \: \: is \: non \: negative \: for \: all \: real \: values \: of \: x \: \: \: and \: \: \frac{7}{4} \: is \: postive }$

$\displaystyle \sf{ \therefore \: { \bigg(x - \frac{5}{2} \bigg)}^{2} \: + \frac{7}{4} \: is \: postive \: for \: all \: real \: values \: of \: x }$

So the required Range is the set of Real numbers

$\sf{Hence \: the \: required \: Range \: = \mathbb{ R}}$

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