Question

x²-7x+10<=0 quadratic inequalities ​

Answer 1

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Answer 2

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

The given quadratic inequality is

$\rm :\longmapsto\: {x}^{2} - 7x + 10 \leqslant 0$

$\rm :\longmapsto\: {x}^{2} - 5x - 2x + 10 \leqslant 0$

$\rm :\longmapsto\:x(x - 5) - 2(x - 5) \leqslant 0$

$\rm :\longmapsto\:(x - 5)(x - 2) \leqslant 0$

We know,

If a and b are positive real numbers such that a < b then

$\boxed{ \sf{ \: (x - a)(x - b) \leqslant 0 \: \implies \: a \leqslant x \leqslant b}}$

So, using this rule

$\bf\implies \:2 \leqslant x \leqslant 5$

$\bf\implies \:x \: \in \: [2, \: 5]$

More to know :-

If a and b are positive real numbers such that a < b, then

$\boxed{ \sf{ \: (x - a)(x - b) < 0 \: \implies \: a < x < b}}$

$\boxed{ \sf{ \: (x - a)(x - b) > 0 \: \implies \: x < a \: or \: x > b}}$

$\boxed{ \sf{ \: (x - a)(x - b) \geqslant0 \: \implies \: x \leqslant a \: or \: x \geqslant b}}$

$\boxed{ \sf{ \: xy > 0 \: \implies \: x > 0,y > 0 \: \: or \: x < 0,y < 0}}$

$\boxed{ \sf{ \: xy < 0 \: \implies \: x > 0,y < 0 \: \: or \: x < 0,y > 0}}$