MHB Show that the solution is not suitable....

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The discussion revolves around demonstrating that the solution to the quadratic equation $5x^2-5x-11=0$, specifically $x=\frac{5-7\sqrt{5}}{10}$, is unsuitable for representing a physical measure. By substituting this value into the expression for the length of PB, calculated as $PB=2x-1$, the result simplifies to a negative value, $\frac{-7\sqrt{5}}{5}$. Negative lengths are not physically meaningful, which confirms the solution's unsuitability. Participants discuss the nature of numbers that can represent physical measures, emphasizing that negative values cannot. The conclusion is that the derived solution does not yield a valid physical length.
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The solution of the quadratic equation $5x^2-5x-11=0$ is $x=\frac{5-7\sqrt{5}}{10}$

PB=$2x-1$ cm

Where do I need help

By substituting the solution $x=\frac{5-7\sqrt{5}}{10}$ in the expression above for the length of $PB$ , show that this solution is not suitable.

Many Thanks :)
 
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What kind of number cannot represent a physical measure?
 
MarkFL said:
What kind of number cannot represent a physical measure?

A fraction or a decimal :)
 
mathlearn said:
A fraction or a decimal :)

No...what's the formula for the distance between two numbers on a number line?
 
MarkFL said:
No...what's the formula for the distance between two numbers on a number line?

(Thinking) I'm sorry i don't know
 
Suppose that $p$ and $q$ are two real numbers. Then the distance $d$ between these numbers, given in units, on a real number line is:

$$d=\sqrt{(p-q)^2}=|p-q|$$

So, what is the range of values we can get for $d$?
 
MarkFL said:
Suppose that $p$ and $q$ are two real numbers. Then the distance $d$ between these numbers, given in units, on a real number line is:

$$d=\sqrt{(p-q)^2}=|p-q|$$

So, what is the range of values we can get for $d$?

My Apologies MarkFL , I don't know that either (Doh)
 
mathlearn said:
MarkFL said:
What kind of number cannot represent a physical measure?
A fraction or a decimal :)

I think we can measure 1/6 of an inch, or 0.12 cm, can't we?
But a length cannot be negative... (Thinking)
 
I like Serena said:
I think we can measure 1/6 of an inch, or 0.12 cm, can't we?
But a length cannot be negative... (Thinking)

Agreed (Nod) Now what is meant above is that as $x$ obtained by simplifying the quadratic equation is negative It is not suitable?...(Thinking)
 
  • #10
mathlearn said:
PB=$2x-1$ cm

By substituting the solution $x=\frac{5-7\sqrt{5}}{10}$ in the expression above for the length of $PB$ ...

After doing that, what do you find?
 
  • #11
greg1313 said:
After doing that, what do you find?

Yeah why not simplify it (Sun)

With PB=$2x-1$ given,

$2*\frac{5-7\sqrt{5}}{10}-1$

$\frac{5-7\sqrt{5}}{5}-1$

$\frac{5-7\sqrt{5}}{5}-\frac{5}{5}$

$\frac{-7\sqrt{5}}{5}$
 

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