Math, asked by briannademmi, 5 months ago

Find a quadratic function with integer coefficients that has zeroes 2/3 and 1/4. Be sure to show all of your work and how the equation will end with 2/3 and 1/4 as the zeroes.

Answers

Answered by nagamanohar21
1

Answer:

I got these solutions by solving the factors. That's "working frontwards". I can also "work backwards" from the solutions. For instance, for x = 2 to be a solution, then I must have solved the factor equation x – 2 = 0, which means that x – 2 must have been a factor.

The factors we find by working backwards from the zeroes are always of the form "(variable) minus (the given zero)". Having a factor of "(variable) minus (value)" means the same thing as having a solution of "(variable) equals (value)"; that is, if "x – a" is a factor, then "x = a" is a solution, and vice versa. We use this fact to find quadratics from their roots.

Find a quadratic with zeroes at 4 and –5.

If the zeroes are at x = 4 and at x = –5, then, subtracting, the factor equations were x – 4 = 0 and x – (–5) = x + 5 = 0. Then the factors were x – 4 and x + 5 . Any factorable quadratic is going to have just the two factors, so these must be them. Then the original quadratic was something like:

(x – 4)(x + 5) = x2 – 4x + 5x – 20 = x2 + x – 20

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Why did I say that the quadratic would be "something like" x2 + x – 20? Because they may have divided something out when they solved the original quadratic. For instance, if I had to solve 2x2 – 2 = 0, I would first divide off the 2 to get x2 – 1 = (x + 1)(x – 1) = 0, with solutions at x = ±1. But if I multiply back only the factors with variables, I get (x – 1)(x + 1) = x2 – 1, which is not quite what I started with. Any number of other quadratics would also share those same two zeroes, including

5x2 – 5, 17x2 – 17, and 4 – 4x2. If they only give me the zeroes, I can't tell if they divided anything out. The quadratic answer I gave in the problem above is good enough, though, because they only asked for "a" quadratic with the given zeroes, not "the" quadratic.

Answered by Anonymous
0

Answer:

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