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slega [8]
4 years ago
8

Find three consecutive positive integers such that the square of the first is equal to the Third

Mathematics
1 answer:
Anna71 [15]4 years ago
5 0
3 consecutive positive integers : x, x + 1, x + 2

x^2 = x + 2
x^2 - x - 2 = 0
(x + 1)(x - 2) = 0

x + 1 = 0
x = -1....this is an extraneous solution that does not work

x - 2 = 0
x = 2

check...
square of the first = third
x^2 = x + 2
2^2 = 2 + 2
4 = 4 (correct)

so ur 3 integers are 2,3, and 4
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PLEASE HELP WILL MARK BRAINLIEST
Sholpan [36]

Answer: Choice D

b greater-than 3 and StartFraction 2 over 15 EndFraction

In other words,

b > 3 & 2/15

or

b > 3\frac{2}{15}\\\\

========================================================

Explanation:

Let's convert the mixed number 2 & 3/5 into an improper fraction.

We'll use the rule

a & b/c = (a*c + b)/c

In this case, a = 2, b = 3, c = 5

So,

a & b/c = (a*c + b)/c

2 & 3/5 = (2*5 + 3)/5

2 & 3/5 = (10 + 3)/5

2 & 3/5 = 13/5

The inequality 2 \frac{3}{5} < b - \frac{8}{15}\\\\ is the same as \frac{13}{5} < b - \frac{8}{15}\\\\

---------------------

Let's multiply both sides by 15 to clear out the fractions

\frac{13}{5} < b - \frac{8}{15}\\\\15*\frac{13}{5} < 15*\left(b - \frac{8}{15}\right)\\\\39 < 15b-8\\\\

---------------------

Now isolate the variable b

39 < 15b-8\\\\15b-8 > 39\\\\15b > 39+8\\\\15b > 47\\\\b > \frac{47}{15}\\\\b > \frac{45+2}{15}\\\\b > \frac{45}{15}+\frac{2}{15}\\\\b > 3+\frac{2}{15}\\\\b > 3\frac{2}{15}\\\\

Side note: Another way to go from 47/15 to 3 & 2/15 is to notice how

47/15 = 3 remainder 2

The 3 is the whole part while 2 helps form the fractional part. The denominator stays at 15 the whole time.

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