find three consecutive even integers such that the product of the first and the second is 92 more than twice the third integer
Answers
Answer:
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Step-by-step explanation:
Three consecutive even integers can be represented by x, x+2, x+4. The sum is 3x+6, which is equal to 108. Thus, 3x+6=108. Solving for x yields x=34.
Answer:
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Step-by-step explanation:
The problem mentions three consecutive even integers we will define as follow:
x = first even integer
x + 2 = second even integer
x + 4 = third even integer
The product of the first and second is 8 more than 6 times the third gives us the following equation:
x(x + 2) = 6(x+4) + 8
x2 + 2x = 6x + 24 + 8 (distributive property on both sides of equation)
x2 + 2x = 6x + 32
x2 - 4x - 32 = 0 (subtract 6x and 32 from both sides of equation)
You can solve a quadratic by factoring or using the quadratic equation.
ax2 + bx + c = 0 is the standard form of a quadratic equation
To factor our quadratic equation we have the following:
a = 1, b = -4, c = -32, a times c = ac = -32