anonymous
  • anonymous
a^2+14a-51=0
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
a=3,-17
anonymous
  • anonymous
how to work it out
mathstudent55
  • mathstudent55
Do you know how to factor the left side?

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anonymous
  • anonymous
like add 51 on both sides
mathstudent55
  • mathstudent55
No, you need to write the left side as the product of two quantities.
anonymous
  • anonymous
how
mathstudent55
  • mathstudent55
Since you have three terms, and one of them has a squared variable, follow these factoring rules: 1. First, try to factor a common factor out. In this case, 1, 14 and -51 have no common factor so there's no common factor to take out. 2. Set up two sets of parentheses: (a )(a )
mathstudent55
  • mathstudent55
Now you are looking for two numbers that when you multiply together you get -51, and when you add together you get 14.
mathstudent55
  • mathstudent55
Can you think of two such numbers?
anonymous
  • anonymous
17 and 3
mathstudent55
  • mathstudent55
Right, but since the product needs to be NEGATIVE 51, and the sum +14, you need +17 and -3. -17 and + 3 will not work because although their product is -51, their sum is -14, not +14.
mathstudent55
  • mathstudent55
Ok, so now you know the numbers are -3 and 17. Just place them inside the two parentheses like this: (a - 3)(x + 17) = 0
mathstudent55
  • mathstudent55
Now the left side is factored.
mathstudent55
  • mathstudent55
The next step is the following: You have a product whose result is zero. How can you get a zero as the result of the multiplication of two numbers?
mathstudent55
  • mathstudent55
The product that is zero is (a + 17) times (a - 3). The only way that product can be zero is if one of the factors is zero. Now you need to see what values of a make these factors zero, so you write: a - 3 = 0 or a + 17 = 0 Solve these simple equations and you get: a = 3 or a = -17
anonymous
  • anonymous
so i can do this on every problem to get dis answer
mathstudent55
  • mathstudent55
This is the method for problems of the type: x^2 + ax + c = 0 where a and b are numbers, and x is the variable. For example: x^2 - x - 2 = 0 y^2 - y - 12 = 0
anonymous
  • anonymous
ok x^2+6x+8=0 answer is -2,-4
mathstudent55
  • mathstudent55
(x + 4)(x + 2) = 0 x + 4 = 0 or x + 2 = 0 x = -4 or x = -2 You are correct.
anonymous
  • anonymous
ok i got it thanks
mathstudent55
  • mathstudent55
wlcm

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