Understanding Quadratic Equations: Solving for x

[headbang]

Can someone please help me understand this.
maybe a solution with explanation ..

$$\frac{2x-1}{x+3}<1$$

Maby the NeXT step..??

$$ 2x-1<x+3$$

Can we use this in the $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

 

[evinda]

Can someone please help me understand this.
maybe a solution with explanation ..

$$\frac{2x-1}{x+3}<1$$

Maby the NeXT step..??
Can we use this in the $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

Hi!

You have to take cases for the denominator.

• $x+3>0 \Rightarrow x>-3$
  
  Then:
  
  $$2x-1<x+3 \Rightarrow x<4$$
• $x+3<0 \Rightarrow x<-3$
  
  $$2x-1>x+3 \Rightarrow x>4, \text{ that is not possible, since , in this case, } x<-3$$
  

So, the only solution is $-3<x<4$.

 

[headbang]

Or maybe:

$$2x-x<3+1$$

$$x<4$$

 

[evinda]

Or maybe:

$$2x-x<3+1$$

$$x<4$$

This stands only if $x+3>0$.

When we have:

$$\frac{a}{b}<1 $$

and we multiply by $b$, we have to distinguish cases.

If $b>0$:
$$\frac{a}{b} \cdot b<1 \cdot b \Rightarrow a<b$$

If $b<0$:

$$\frac{a}{b}<1 \Rightarrow a>b$$

For example:

$$\frac{5}{-3}<1 \Rightarrow \frac{5}{-3} \cdot (-3)> 1 \cdot (-3) \Rightarrow 5>-3$$

Multiplying both sides of an inequality by a negative number, the direction of the inequality changes ("<" becomes ">").

 

[MarkFL]

Can someone please help me understand this.
maybe a solution with explanation ..

$$\frac{2x-1}{x+3}<1$$

Maby the NeXT step..??

$$ 2x-1<x+3$$

Can we use this in the $$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$

As mentioned, you don't want to multiply through by a quantity containing the variable, because it may be negative , positive or zero. I would subtract 1 from bot sides to get:

$$\frac{2x-1}{x+3}-1<0$$

Combine terms on the left:

$$\frac{2x-1-(x+3)}{x+3}<0$$

$$\frac{2x-1-x-3}{x+3}<0$$

$$\frac{x-4}{x+3}<0$$

Okay, we now know that the sign of the expression on the left will change at two critical values:

$$x=-3,\,4$$

Because the roots in the numerator and denominator occur an odd number of times (in this case both once), we know the sign of the expression will alternate across the 3 intervals of the real number line into these two critical values divide it. So, I would test a number in the middle interval, specifically $x=0$, and we see that the expression is negative, and satisfies the inequality. Thus, in the left and right intervals, the expression will be positive, and not part of the solution. Since the inequality is strict, the interval is open, thus the solution, in interval notation is:

$$(-3,4)$$