The roots of x^2-6x+c=0 differ by 2 squareroot of 3.?

Question

a. What are the roots of the equation?
B. What is the value of c?

Answer 1

We have the equation
x² - 6x + c = 0

Now, we let p and q be the roots of the equation. If we are given the quadratic equation ax² + bx + c = 0, then the sum of the roots is -b/a. Thus,
p + q = -b/a = -(-6)/(1) = 6
p + q = 6

Now, since they differ by 2√3, then
p - q = 2√3

Thus, we now have 2 equations:
p + q = 6
p - q = 2√3

We now add them
2p = 6 + 2√3

Thus,
p = 3 + √3

We can solve for q:
p + q = 6
q = 6 - p
q = 6 - (3 + √3)
q = 3 - √3

Therefore, the roots of the equation are:
3 + √3 and 3 - √3

Now, given the quadratic ax² + bx + c = 0, the product of the roots is c/a. Thus,
pq = c/a = c/1 = c

Now,
pq = c

We can substitute the value of p and q
(3 + √3)(3 - √3) = c

thus, using the special multiplication (a + b)(a - b) = a² - b²
c = 3² - √3²

Thus,
c = 9 - 3

therefore,
c = 6

Now, here are the answers to your question:
a) What are the roots of the equation?
3 + √3 and 3 - √3
B) what is the value of c?
c = 6

Hope this helped!

^_^

Answer 2

x² -6x + c = 0
The roots are a number (n) that differ by 2√3.
→ x = n + 2√3 and x = n
(x - n - 2√3)(x - n) = 0
When you cross multiply, the result will equal to the b term.
Multiply the x of the 1st. factor with the number of the second factor,
and multiply the number of the 1st factor with the number of the 2nd factor, and add the results.

-nx - 2√3x -nx = -6x
-2nx - 2√3x = -6x (÷ -2)
nx + √3x = 3x
x(n + √3) = 3x
n + √3 = 3x/x
n + √3 = 3
n = 3 - √3
n = 1∙267 949 192

Now, get the value of the roots.
x = n + 2√3 and x = n
x = 1∙267 949 192 + 2√3 and x = 1∙267 949 192
x = 4∙732 050 808 and x = 1∙267 949 192

Now, use the x values to calculate the equation.
x = 4∙732 050 808 and x = 1∙267 949 192
(x - 4∙732 050 808)(x - 1∙267 949 192) = 0
x² - 1∙267 949 192x - 4∙732 050 808x + 6 = 0
x² - 6x + 6 = 0

→ Roots are : x = 4∙732 050 808 and x = 1∙267 949 192
→ Roots are : x = 3 + √3 and x = 3 - √3

→ c = +6.

Answer 3

the root differ by 2 sqrt(3)

let them be a+ sqrt(3) and a- sqrt(3)

sum of roots 2a = 6
a = 3
roots are 3+ sqrt(3) and 3-sqrt(3)

c = product of roots = 3^2-3 =6

so c = 6