Word Problems(might involve factoring)

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SUMMARY

This discussion focuses on solving various word problems involving algebraic equations and factoring techniques. Key problems include finding a number based on a quadratic equation, determining consecutive integers whose product equals a specific value, and solving for consecutive positive odd integers. The participants demonstrate the use of factoring methods, such as transforming equations into standard form and applying the quadratic formula. The final problem involves finding two consecutive integers whose squares sum to 41, leading to the solutions (4, 5) and (-4, -5).

PREREQUISITES
  • Understanding of quadratic equations and their standard forms
  • Familiarity with factoring techniques in algebra
  • Knowledge of solving word problems using algebraic expressions
  • Basic skills in manipulating and simplifying algebraic equations
NEXT STEPS
  • Study the quadratic formula and its applications in solving equations
  • Practice factoring polynomials, specifically quadratic expressions
  • Explore techniques for solving word problems in algebra
  • Learn about the properties of consecutive integers and their applications in problem-solving
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Students learning algebra, educators teaching mathematical problem-solving, and anyone interested in enhancing their skills in factoring and solving equations.

Hardeep
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Im having a bit of trouble with these:

1). Seven less than 4 times the square of a number is 18. Find the number.

2). Find two consecutive integers whose product is 56.

3). Find two consecutive positive odd integers whose product is 35.
 
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4n^2 - 7 = 18
4n^2 = 25
n^2=6.25
n=2.5
 
n x (n+1) = 56
its 7 and 8 .. but I can't seem to remember how to get the answer with the equation

nx (n+2) = 35

5 and 7 =)
 
Help_Me_Please said:
4n^2 - 7 = 18
4n^2 = 25
n^2=6.25
n=2.5

I know that why, but we have to factor to get it. Like

x^2 - x = 6
x^2 - x - 6 = 0
(x + 2)(x -3)

x= -2 OR x= 3
 
4n^2 - 25 = 0
4n^2 -10n + 10 n -25 = 0
2n (4n^2 - 10n) + ( 10n - 25) 5
2n - 5 + 2n- 5 2n+5

2n-5 and 2n+5 =)
 
what I did was I took the 25 and multiplied it by the 4 and got 100 ... then I thought of factors of 100 that would add to give me 0 .. and just put it in like it was part of the problem
 
Help_Me_Please said:
what I did was I took the 25 and multiplied it by the 4 and got 100 ... then I thought of factors of 100 that would add to give me 0 .. and just put it in like it was part of the problem

I see now, thanks fo rthe help
 
Last edited:
ok I have just one more;

4). The sum of the sqaures of two consective integers is 41. FInd the integers.
 
Hardeep said:
ok I have just one more;

4). The sum of the squares of two consective integers is 41. FInd the integers.

Call them 'a' and 'b'.Then
a^{2}+b^{2}=41.Since 'a','b' are integers,then both 'a' and 'b' in modulus must be smaller or at most equal to 6,as 6 is the greatest integer whose square is less/equal to 41.
The only possible sollutions to the eq. are (4,5),(-4,-5) and viceversa.The mixed combinations don't have consecutive integers.

Daniel.
 

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