Tricky equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

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In summary, the conversation discusses the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w and how it can be simplified to a quadratic equation in w^2. The solution to this equation can be used to find the area of a triangle using Heron's formula. The conversation also includes an example and a question about why one particular solution does not work.
  • #1
Wilmer
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a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

Given a, u and v, w = ?
 
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  • #2
Wilmer said:
a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

Given a, u and v, w = ?

\(\displaystyle k = \dfrac{u+v+w}{2}\), \(\displaystyle k-u = \dfrac{-u+v+w}{2}\), \(\displaystyle k-v = \dfrac{u-v+w}{2}\), \(\displaystyle k-w = \dfrac{u+v-w}{2}\).

Substituting in we get

\(\displaystyle a^2=\dfrac{1}{16}(u+v+w)(u+v-w)(w+(u-v))(w-(u-v))\)

Simplifying yields

\(\displaystyle 16a^2=((u+v)^2-w^2)(w^2-(u-v)^2)\)

\(\displaystyle -16a^2=(w^2-(u+v)^2)(w^2-(u-v)^2)\)

\(\displaystyle w^4-2(u^2+v^2)w^2+(u^2-v^2)^2+16a^2=0\)

This is a quadratic equation in \(\displaystyle w^2\). Substitute in the values for \(\displaystyle a,u,v\) and use the quadratic formula.
 
  • #3
I was able to get to:
w = SQRT[u^2 + v^2 + 2SQRT((uv)^2 - 4a^2)]

As you probably surmised, this is Heron's triangle area in disguise!

Example: triangle sides u,v,w: u=4, v=13, w=15 : a = area = 24
My solution will give correctly w = 15 using u=4:v=13 or u=13:v=4
But not if u=4, v=15: does not yield 13

Can you tell me why...thanks in advance...
 
  • #4
Wilmer said:
I was able to get to:
w = SQRT[u^2 + v^2 + 2SQRT((uv)^2 - 4a^2)]

As you probably surmised, this is Heron's triangle area in disguise!

Example: triangle sides u,v,w: u=4, v=13, w=15 : a = area = 24
My solution will give correctly w = 15 using u=4:v=13 or u=13:v=4
But not if u=4, v=15: does not yield 13

Can you tell me why...thanks in advance...

Double check your arithmetic.
With \(\displaystyle u=4\), \(\displaystyle v=15\) and \(\displaystyle a=24\), you should get the equation

\(\displaystyle w^4-482w^2+52897=0\) which factors into

\(\displaystyle (w-13)(w+13)(w^2-313)=0\)

This appears to yield two possible answers \(\displaystyle w=13\) and \(\displaystyle w=\sqrt{313}\).
 
  • #5
Gotcha! Thanks again...
 

Related to Tricky equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w

1. What is the significance of the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w?

The equation represents a quadratic relationship between the variable a and the variables k, u, v, and w. The value of k is determined by the sum of u, v, and w, which can provide insight into the relationship between the variables.

2. How do you solve the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w?

To solve the equation, you can use the quadratic formula or factor the equation into two binomials. Once the equation is in factored form, you can set each factor equal to zero and solve for the possible values of a, k, u, v, and w.

3. What does the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w represent in real life?

The equation can represent various real-life situations, such as the relationship between the area of a square (a^2) and the length of its side (k). The variables u, v, and w can represent different quantities that contribute to the overall length of the side.

4. Can you provide an example of a problem that can be solved using the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w?

One example is calculating the area of a trapezoid when given the length of the two parallel sides (u and v) and the length of the non-parallel sides (w). The equation can also be used to find the length of a side of a square when given the area and the value of k.

5. How does the equation a^2 = k(k-u)(k-v)(k-w) where 2k = u+v+w relate to other mathematical concepts?

The equation is related to the concept of factoring and solving quadratic equations. It also involves the use of variables and their relationships to each other. In addition, the equation can be used in geometry to find the area or length of a shape.

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