# Thread: Another Problem - split from "completing square"

1. ## Another Problem - split from "completing square"

alright i have solve it.

but here is another question which i also don't know how to do.
can help me.

Solve the given equations, determine the character of the roots.

In finding the radius r of a circular arch of height h and span b, an architect used the following formula. solve for h

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$

2. Originally Posted by danielwu
alright i have solve it.

but here is another question which i also don't know how to do.
can help me.

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$
We need a bit of a clue as to what you want done with this.

RonL

3. okay i have re-write my post. thanks alot

4. Originally Posted by danielwu
alright i have solve it.

but here is another question which i also don't know how to do.
can help me.

Solve the given equations, determine the character of the roots.

In finding the radius r of a circular arch of height h and span b, an architect used the following formula. solve for h

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$
Thus,
$\displaystyle 8rh=b^2+4h^2$
Thus,
$\displaystyle 4h^2-8rh+b^2=0$
The discriminant is,
$\displaystyle \sqrt{64r^2-16b^2}$
Is it positive?
Are you sure that $\displaystyle r>b$?
If thus, then your two roots are real.

5. Originally Posted by ImPerfectHacker
Are you sure that $\displaystyle r>b$?
If thus, then your two roots are real.
For a circular arch of radius $\displaystyle r$spanning $\displaystyle b$, we must have:

$\displaystyle b \le 2r$

Its simple geometry.

RonL

6. Originally Posted by CaptainBlack
For a circular arch of radius $\displaystyle r$spanning $\displaystyle b$, we must have:

$\displaystyle b \le 2r$

Its simple geometry.

RonL
what is b here? I don't understand

Keep Smiling
Malay

7. Originally Posted by malaygoel
what is b here? I don't understand

Keep Smiling
Malay
It is the gap spanned by the arch.

Think of the arch as forming an arc of a circle, then b is the length of the
corresponding chord.

RonL

8. Originally Posted by CaptainBlack
It is the gap spanned by the arch.

Think of the arch as forming an arc of a circle, then b is the length of the
corresponding chord.

RonL
thanks

Keep Smiling
Malay

9. Originally Posted by ThePerfectHacker
Thus,
$\displaystyle 8rh=b^2+4h^2$
Thus,
$\displaystyle 4h^2-8rh+b^2=0$
The discriminant is,
$\displaystyle \sqrt{64r^2-16b^2}$
Is it positive?
Are you sure that $\displaystyle r>b$?
If thus, then your two roots are real.

hi i am not sure if this is the right answer?
what i have learn in my current chapter for this question is to make use this formula

$\displaystyle x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}$

10. Originally Posted by danielwu
hi i am not sure if this is the right answer?
what i have learn in my current chapter for this question is to make use this formula

$\displaystyle x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}$
I doubt that, you may have been shown:

$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,

and the term $\displaystyle b^2 - 4ac$ under the root sign is the
discriminant that other poster have been referring to.

RonL

11. Originally Posted by danielwu
hi i am not sure if this is the right answer?
what i have learn in my current chapter for this question is to make use this formula

$\displaystyle x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}$
You mean,
$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
---
That is what I did.
You only look at the expression under the radical.
$\displaystyle b^2-4ac$
This tells you its signs.
If it is negative then you have imaginaries.
If it is zero then you have one a singal solution to quadradic.
It it is positive then you have two distint solution.
And if the positive term is a perfect square then you have rational solutions.
To summerize,
$\displaystyle b^2-4ac\left\{ \begin{array}{cc}<0&\mbox{ imaginary}\\ =0&\mbox{a single rational solution}\\ >0&\left\{ \begin{array}{cc}\mbox{square}&\mbox{rational}\\ \mbox{not a square}&\mbox{irrational} \end{array} \end{array} \right$

12. Originally Posted by ThePerfectHacker
Thus,
$\displaystyle 8rh=b^2+4h^2$
Thus,
$\displaystyle 4h^2-8rh+b^2=0$
The discriminant is,
$\displaystyle \sqrt{64r^2-16b^2}$
Is it positive?
Are you sure that $\displaystyle r>b$?
If thus, then your two roots are real.

hi i am not sure if this is the right answer?
what i have learn in my current chapter for this question is to make use this formula

$\displaystyle x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}$

Example:

If a = 3, b= -3, c = -2 for $\displaystyle 3y^2 - 3y -2 = 0$

Then what is for

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$
a = ?, b = ?, c = ?

13. Originally Posted by ThePerfectHacker

..The discriminant is,
$\displaystyle \sqrt{64r^2-16b^2}$
The discriminant is $\displaystyle {64r^2-16b^2}$

RonL

14. Originally Posted by ThePerfectHacker
Thus,
$\displaystyle 8rh=b^2+4h^2$
Thus,
$\displaystyle 4h^2-8rh+b^2=0$
The discriminant is,
$\displaystyle \sqrt{64r^2-16b^2}$
Is it positive?
Are you sure that $\displaystyle r>b$?
If thus, then your two roots are real.

hi i am not sure if this is the right answer?
what i have learn in my current chapter for this question is to make use this formula

$\displaystyle x = -b \pm \sqrt \frac{b^2 - 4ac}{2a}$

Example:

If a = 3, b= -3, c = -2 for
$\displaystyle 3y^2 - 3y -2 = 0$

Then what is for

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$
a = ?, b = ?, c = ?

15. Originally Posted by danielwu

Then what is for

$\displaystyle r = \frac {b^2 + 4h^2}{8h}$
a = ?, b = ?, c = ?
Multiply by 8h to get,
$\displaystyle 8hr=b^2+4h^2$
Thus,
$\displaystyle 4h^2-8hr+b^2=0$
Thus,
$\displaystyle a=5,b=-8h,c=b^2$

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