Line meets curve

<ol class="abc-list"><li><div class="list-item-paragraph">Given that $x$ is real, prove that ${{x}^{2}}-4x+9$ is always positive.</div><div class="list-item-paragraph">Solve the inequality $\frac{{{x}^{3}}+2{{x}^{2}}+x+14}{{{x}^{2}}+5}>x+1$.</div></li></ol>

Given that $x$ is real, prove that ${{x}^{2}}-4x+9$ is always positive.

Solve the inequality $\frac{{{x}^{3}}+2{{x}^{2}}+x+14}{{{x}^{2}}+5}>x+1$.

Default solution

Since $(x - 2)^2 \geq 0$, $(x - 2)^2 + 5 > 0$

$\therefore x^2 - 4x + 9 > 0$ for all $x \in \mathbb{R}$ (shown)

$\times (x^2 + 5)$: $\frac{x^3 + 2x^2 + x + 14}{x^2 + 5} > x + 1$ since $x^2 + 5 > 0$

$x^3 + 2x^2 + x + 14 > (x + 1)(x^2 + 5)$

$x^3 + 2x^2 + x + 14 > x^3 + 5x + x^2 + 5$

$\therefore$ Since $x^2 - 4x + 9$ is always positive, Solution set $= \{x \in \mathbb{R}\}$

N1998 P1 Q15

N1998 P1 Q15

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Default solution