Adolfo Hebert

2022-03-31

Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed

${x}^{2}-|5x-3|-x<2$

For this problem I took two cases where in one case$5x-3\ge 0$ and in second case $5x-3<0$

1)$5x-3\ge 0$ :

2)$5x-3<0$

For this problem I took two cases where in one case

1)

2)

Abdullah Avery

Beginner2022-04-01Added 19 answers

Let me forcus on when $5x-3<0$, we have $x<\frac{3}{5}$ and ${x}^{2}-|5x-3|-x<2$

Becomes ${x}^{2}-(3-5x)-x<2$

${x}^{2}+4x-5<0$

$(x+5)(x-1)<0$

Hence we have $-5<x<1\text{}\text{and}\text{}x\frac{3}{5}$, hence taking intersection, we have $-5<x<\frac{3}{5}$.

To answer the question, why did we exclude $x=-5$, consider $(x+5)(x-1)<0$, if we substute $x=5$ inside, we get $0<0$ which is not true.

Esteban Sloan

Beginner2022-04-02Added 21 answers

If $5x-3\ge 0$, we have $x\ge \frac{3}{5}$ and

${x}^{2}-|5x-3|-x<2$

becomes ${x}^{2}-5x+3-x<2$

${x}^{2}-6x+1<0$

$\frac{6-\sqrt{36-4}}{2}<x<\frac{6+\sqrt{36-4}}{2}$

$3-2\sqrt{2}<x<3+2\sqrt{2}$

Taking intersection with $x\ge \frac{3}{5}$, we have

$\frac{3}{5}\le x<3+2\sqrt{2}$

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V=??

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a) $f(x)=-3x+4$

b) $f\left(x\right)=-3{x}^{2}+7$

c) $f(x)=\frac{x+1}{x+2}$

?

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A)Rhombus;

B)Parallelogram;

C)Kite;

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$B=1.4\times {10}^{-1}$,

$C=2\times {10}^{3}$,

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A)5;

B)4;

C)6;

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