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Andru [333]
2 years ago
15

Consider the quadratic equation ax2 + bx +5 = 0, where a,b, and c are rational numbers and the quadratic has two disticnt zeros.

Mathematics
1 answer:
Afina-wow [57]2 years ago
4 0

Step-by-step explanation:

x2 + bx +5 = 0,

where a,b and c are rational numbers.

Also it is given that the quadratic has 2 distinct zeros and one of it's zero is irrational.

We know that if it has a one irrational zero, the conjugated couple of the one zero is also a null with the same polynomial balance for a polynomial equation with rational coefficients." Therefore, in this case it is a quadratic equation, one is null and irrational; thus, the other zero is irrational.

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nadya68 [22]

Answer:

(t, t -1, t)

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You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

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\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

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From the first equation in the original system,

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