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a_sh-v [17]
3 years ago
14

Consider the equation 5x^2-10x+c=0. What values of c result in the equation having a complex solutions? Represent your answer on

the number line

Mathematics
1 answer:
stiv31 [10]3 years ago
8 0

Answer:

When we have a quadratic equation:

a*x^2 + b*x + c = 0

There is something called the determinant, and this is:

D = b^2 - 4*a*c

If D < 0, then the we will have complex solutions.

In our case, we have

5*x^2 - 10*x + c = 0

Then the determinant is:

D = (-10)^2 - 4*5*c = 100 - 4*5*c

And we want this to be smaller than zero, then let's find the value of c such that the determinant is exactly zero:

D = 0 = 100 - 4*5*c

    4*5*c = 100  

       20*c = 100

            c = 100/20 = 5

As c is multiplicating the negative term in the equation, if c increases, then we will have that D < 0.

This means that c must be larger than 5 if we want to have complex solutions,

c > 5.

I can not represent this in your number line, but this would be represented with a white dot in the five, that extends infinitely to the right, something like the image below:

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The question to be solved is the following :

Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if b \neq 0. Recall that given two vectors a,b  a⊥ b if and only if a\cdot b =0 where \cdot is the dot product defined in \mathbb{R}^n. Suposse that b\neq 0. We want to find γ such that (a-\gamma b)\cdot b=0. Given that the dot product can be distributed and that it is linear, the following equation is obtained

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