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 #2
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Absolutely, let's find the value of k for the quadratic equation where one solution is twice the other solution.

 

We can approach this problem by letting the solutions be x1​ and x2​ and setting up the relationship between them along with the quadratic equation. Then, we can use the quadratic formula to solve for k.

 

Steps to solve: 1. Set up the relationships: Let x1​ and x2​ be the solutions of the quadratic equation. We are given that one solution is twice the other, so we can represent this as:

 

x1​=2x2​

We can substitute this into the quadratic equation:

 

x2+12x+k=0

 

This becomes: (2x2​)2+12(2x2​)+k=0

 

Expanding the equation: 4x22​+24x2​+k=0

 

2. Solve for k using the quadratic formula:

 

Since we know the relationship between x1​ and x2​, we can rewrite the equation in terms of x2​ only. Factoring out x2​, we get:

 

x2​(4x2​+24+k)=0

 

For the equation to hold true, either x2​ must be zero (which would mean both solutions are zero) or the expression in the parenthesis must equal zero.

 

Since we're given that there are two distinct solutions, we know x2​ cannot be zero. Therefore:

 

4x2​+24+k=0

 

This is a quadratic equation in terms of x2​. We can solve for k using the quadratic formula, which is:

 

k=(−b±b2−4ac​)/2a

 

In this case, a = 4, b = 24, and c = k. Substituting these values into the formula, we get:

 

k=(−24±242−4∗4∗k​)/2∗4

 

3. Solve for k:

 

Since we are given that there is only one solution (one root), the discriminant (the expression under the square root) must be zero. Setting the discriminant to zero and solving for k, we get:

 

242−4∗4∗k=0 576=16k k=16576​ k=−36

 

Answer: The value of k for the quadratic equation where one solution is twice the other solution is k = -36.

Mar 19, 2024
 #1
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Consider the expression (3​/2−i/2)n. Let's analyze the powers of this complex number according to the four cases for the remainder r of n when divided by 4:

 

Case r=0: Then n=4k for some integer k, and

 

[(\sqrt{3}/2 - i/2)^{4k} = \left( (\sqrt{3}/2)^2 + (i/2)^2 \right)^{2k} = \left( \frac{3}{4} + \frac{1}{4} \right)^{2k} = 1^{2k} = 1.]

 

Since 1 is real, this case produces a real number.

 

Case r=1: Then n=4k+1 for some integer k, and

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 1} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= (\sqrt{3}/2 - i/2) \cdot 1 \ &= \sqrt{3}/2 - i/2. \end{align*}

 

Since this expression has a nonzero imaginary part, it is not real.

 

Case r=2: Then n=4k+2 for some integer k, and

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 2} &= (\sqrt{3}/2 - i/2)^2 \cdot (\sqrt{3}/2 - i/2)^{4k} \ &= \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i - \frac{1}{4} \right) \cdot 1 \ &= \frac{3}{4} - \frac{2 \sqrt{3}}{4} i. \end{align*}

 

Since this expression has a nonzero imaginary part, it is not real.

 

Case r=3: Then n=4k+3 for some integer k, and

 

\begin{align*} (\sqrt{3}/2 - i/2)^{4k + 3} &= (\sqrt{3}/2 - i/2) \cdot (\sqrt{3}/2 - i/2)^{4k + 2} \ &= (\sqrt{3}/2 - i/2) \cdot \left( \frac{3}{4} - \frac{2 \sqrt{3}}{4} i \right) \ &= \frac{3 \sqrt{3}}{4} + \frac{1}{4} - \frac{3}{4} i. \end{align*}

 

Since this expression has a nonzero imaginary part, it is not real.

 

In conclusion, the smallest positive integer value of n such that (3​/2−i/2)^n is real is 4​.

Mar 17, 2024