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Radda [10]
3 years ago
15

find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second

Mathematics
1 answer:
melamori03 [73]3 years ago
5 0

Answer:

Step-by-step explanation:

We are looking for two numbers, N and N + 1. From the problem statement, we can setup the following equation:

3N^{2} = 5(N + 1) + 7

3N^{2} = 5N + 5 + 7

3N^{2} - 5N - 12 = 0

(3N + 4)(N - 3)

N = \frac{-4}{3}, 3

Because 3 is the only integer solution, the answer is 3 and 4.

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<span>First, we figure out the value of c or the y intercept, we use the second point (0, 1) and substitute to the equation of the parabola. W</span><span>hen x = 0, y = 1. So, c should be equal to 1. The</span><span> parabola is y = ax^2 + bx + 1 </span>

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<span>The slope at x = 1 is equal to 4 which is equal to the first derivative of the equation.</span>
<span>We take the derivative of the  equation ,
y = ax^2 + bx + 1</span>
<span>y' = 2ax + b 
</span>
<span>x = 1, y' = 2
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<span>4 = 2a + b </span>

So, we have two equations and  two unknowns,<span> </span>
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<span>b = -6</span>

<span>Therefore, the eqution of the parabola is y = 5x^2 - 6x + 1 .</span>
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