Find the product of all real values of $r$ for which $\frac{1}{2x}=\frac{r-x}{2}$ has exactly one real solution.

maximum Aug 7, 2023

#1**+1 **

*Note: This sounds suspiciously like an Alcumus problem that I've done before.*

**This is just a quadratic. \(\frac{1}{2x} = \frac{r-x}{2}\)**

**So, \(2 = 2xr-2x^2\)**

**\(2x^2-2xr+2 = 0\)**

**Now just plug in the quadratic formula and you should get the answer, don't forget to multiply.**

**P.S I left some of the problem for you to do, just to make sure that you are not cheating**

**Hope this helped, EJ**

**Why my posts always flag for moderation Cphil?**

Imcool Aug 7, 2023

#2**+1 **

1/ (2x) = ( r - x) / 2 cross-multiply

2 = 2x ( r - x)

2 = 2xr -2x^2 rearrange as

2x^2 -2xr + 2 = 0 simplify

x^2 - xr + 1 = 0

This will have one real solution when

r^2 - 4(1)(1) = 0

r^2 - 4 = 0

r^2 = 4 take both roots

r = 2 , -2

And the product of these = -4

CPhill Aug 7, 2023

#7**0 **

We can solve this problem using the discriminant of the quadratic equation. For a quadratic equation of the form ax^2 + bx + c = 0, the discriminant is b^2 - 4ac. If the discriminant is positive, the quadratic equation has two distinct real roots. If the discriminant is zero, the quadratic equation has one real root. If the discriminant is negative, the quadratic equation has no real roots.

In this problem, we are given the quadratic equation x^2 - (a + 1)x + a = 0, and we want to find the product of all values of a for which the equation has exactly one real solution.

Using the quadratic formula, we can solve for x:

x = [(a + 1) ± sqrt((a + 1)^2 - 4a)] / 2

The equation has exactly one real solution when the discriminant is zero:

(a + 1)^2 - 4a = 0

Simplifying this equation, we get:

a^2 - 2a + 1 = 0

This equation has a double root at a = 1. Therefore, the product of all values of a for which the equation has exactly one real solution is 1.

lowergm Aug 7, 2023