Here's the question you clicked on:
zammett.damali
a^2+14a-51=0
how to work it out
Do you know how to factor the left side?
like add 51 on both sides
No, you need to write the left side as the product of two quantities.
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 )
Now you are looking for two numbers that when you multiply together you get -51, and when you add together you get 14.
Can you think of two such numbers?
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.
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
Now the left side is factored.
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?
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
so i can do this on every problem to get dis answer
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
ok x^2+6x+8=0 answer is -2,-4
(x + 4)(x + 2) = 0 x + 4 = 0 or x + 2 = 0 x = -4 or x = -2 You are correct.
ok i got it thanks