Out of the three consecutive positive integers ,the middle number is p.If three times the square of the largest is greater than the sum of the other two numbers by 67;calculate the value of p @mathslover @sepeario
Let's label our three numbers a, b, and c (in ascending order). From the details you provided: \(3c^2=67+a+b\) We also know that they are consecutive and positive, so: \(a+1=b; b+1=c \) Let's define all of this in terms of b: \(a=b-1 ; ~ c=b+1\) now, we want to rewrite the equation we are given in terms of b. \(3(b+1)^2=b+(b-1)+67\) Does it make sense so far?
No
So we have three consecutive positive numbers. Like, 7 8 9. The middle one is labeled "p", so we have \(p-1, ~~ p, ~~ p+1\) as our three consecutive integers. "Three times the square of the largest" \(\rightarrow 3(p+1)^2\) "Is greater than...by 67" \(\rightarrow =67+ \dots\) "The sum of the other two numbers" \(\rightarrow (p-1)+p\) So, altogether, we have: \(3(p+1)^2=67+(p-1)+p\) When we expand that out, we get: \(3(p^2+2p+1)=2p+67-1 \rightarrow 3p^2+6p+3=2p+66\) \(\rightarrow 3p^2+4p-63=0\) You should be able to factor that last part and solve for p.
p is the middle number p-1 is the number before p and p +1 means the number after p if you didn't know
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@Sepeario
Do you know how to factor polynomials like the one given above?
Look at this website it has the exact same question :P http://www.algebra.com/algebra/homework/Problems-with-consecutive-odd-even-integers/Problems-with-consecutive-odd-even-integers.faq.question.916180.html
@BTaylor you are on the right track, but your polynomial is incorrect and hence cannot be factorised.
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Did you look at the website?
THANKS MATE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Can u help me more?
I'll try. I would also like to mention that every question you've asked involves a quadratic equation and/or the quadratic formula, so you would do good by learning those well.
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