How do you solve $4 x^{2} < (12 x - 9)$ $4x^2 <(12x-9)$ ?

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Answer 1 · 2016-09-02T02:50:34

$4 x^{2} < 12 x - 9$ $4x^2<12x-9$ does not have any solution.

$4 x^{2} < 12 x - 9$ $4x^2<12x-9$

$\Leftrightarrow 4 x^{2} - 12 x + 9 < 0$ $hArr4x^2-12x+9<0$ or

${(2 x)}^{2} - 2 \times 2 x \times 3 + 3^{2} < 0$ $(2x)^2-2×2x×3+3^2<0$ or

${(2 x - 3)}^{2} < 0$ $(2x-3)^2<0$ i.e. $(2 x - 3$ $(2x-3$ is negative.

But a square number can not be negative.

Hence $4 x^{2} < 12 x - 9$ $4x^2<12x-9$ does not have any solution.

How do you solve 4x^2 <(12x-9)4x2<(12x−9)4x^2 <(12x-9)?