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Vesna [10]
2 years ago
8

What are the solutions to 3x^2 = 8x — 6?

Mathematics
1 answer:
Bond [772]2 years ago
5 0

Answer:

First, let's move all terms to one side,

\sf x^2-8x+6=0

When given this form (ax^2+bx+c=0), there will be two solutions at the end.

1) \sf x=\frac{-b+\sqrt{b^2-4ab} }{2b}

2) \sf x=\frac{-b-\sqrt{b^2-4ab} }{2b}

where, a=1, b=-8, c=6

\sf x=\frac{8+\sqrt{(-8)^2-4\times6} }{2} ,\frac{8-\sqrt{(-8)^2-4\times6} }{2}

Simplify,

\sf x=\frac{8+2\sqrt{10} }{2} ,\frac{8-2\sqrt{10} }{2}

Simplify,

\sf x=4+\sqrt{10}, 4-\sqrt{10}

Hope it helps!

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Answer:

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For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

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There isn't a b variable in this equation, but x = 2.5. (:

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