Question

Hey guys

Solve the following inequality

1) x^2-7x<0

2)x^2-3x-1 > 0

3) x^2-7x-1 < _ 0

4) 4-x^2 >_ 0

5) 9x-x^2 > _ 0

If u know then answer otherwise leave it.

Answer 1

Hai "here is your answer.

Answer 2

Hey friend, Harish here.

Here is your answer.

$1) \ x^2 - 7x \ \textless \ 0 \\ \\ \rightarrow x^2 \ \textless \ 7x \\ \\ \to x\ \textless \ 7, and \ x\ must\ be\ greater\ than\ zero \\ \\ \boxed{\bold{Therefore, 0\ \textless \ x \ \textless \ 7 }} \\ \\ \\ 2) \ x^2 -3x -1 \ \textgreater \ 0 \\ \\ \to \left(x-\frac{3+\sqrt{13}}{2}\right)\left(x-\frac{3-\sqrt{13}}{2}\right) \ \textgreater \ 0 \\ \\\boxed{\bold{Therefore\ \:x\ \textless \ \frac{3-\sqrt{13}}{2}\quad \mathrm{or}\quad \:x\ \textgreater \ \frac{3+\sqrt{13}}{2} }} \\ \\ \\ 3) \ x^2-7x-1\:\le 0 \\ \\ \to \left(x-\frac{7+\sqrt{53}}{2}\right)\left(x-\frac{7-\sqrt{53}}{2}\right)\le \:0$ 

$\boxed{\bold{Therefore, \frac{7-\sqrt{53}}{2}\le \:x\le \frac{7+\sqrt{53}}{2}}} \\ \\ \\ 4) \ 4-x^2 \ge 0 \\ \\ \to 4 \ge x^2 \\ \\ \to x\le \:2\quad \mathrm{and}\quad \left(x\ge \:-2\right) \\ \\ \boxed{\bold{Therefore, -2\le \:x\le \:2}} \\ \\ \\ 5) \ 9x - x^2 \ge 0 \\ \\ \to By\ inspection, \ we \ can\ see\ that\ 0\ satisfies\ this\ inequality. \\ \\ And, \\ \\ \to 9x \ge x^2 \\ \\ \to 9 \ge x \\ \\ \boxed{\bold{Therefore, 0\le \:x\le \:9}}$
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Hope my answer is helpful to you.