The equation b^2 + 10b = 39 is similar to the 'Quadratic Equation:'
y = ax^2 + bx + c
Instead of the term x, you have replaced it with the term 'b.'
So, in theory, to find x (in this case b) we make y = 0 by inversing 39 from the right hand side (RHS) to the left hand side (LHS):
b^2 + 10b - 39 = 0
Notice the similarity between this equation and the original quadratic equation:
b^2 + 10b - 39 = 0 ax^2 + bx + c = y
The term a = 1 for your formula, b = 10, and c = -39. The term x = b = ?, and y = 0.
*From now on, I'll use the original quadratic formula terms. So your formula will look like this:
x^2 + 10x - 39 = 0
We can find the solution to x through two methods: sum and product, or the quadratic formula.
METHOD ONE: SUM AND PRODUCT To find x, we find the sum and product of a × c, and b.
a × c = 1 × -39 = -39
b = 10
Now we need two numbers that have a sum of 10 and a product of -39.
The two numbers are 13 and -3. Proof:
13 × -3 = -39 {which is the product}
13 + -3 = 13 -3 {a + and - make a -} = 10 {which is the sum}
Now that we know these are the two numbers, we rewrite the equation like so:
(x + 13) (x - 3) = 0
Now, to find x, we know that either (x + 13) or (x - 3) equals 0, because the value x are the x-intercepts of the graph. And the x-intercepts are where y = 0. So:
x + 13 = 0
OR
x - 3 = 0
Now we inverse the values that we do know from the LHS to the RHS (see above for definitions of LHS and RHS):
x = -13
OR
x = 3
So the values of x (or the x-intercepts) are -13 and 3. So for your case b = -13 and 3.
We can prove this by replacing x in the formula:
x^2 + 10x - 39 = (-13)^2 + 10(-13) - 39 = 169 + (-130) - 39 = 169 - 130 - 39 = 0 {therefore -13 is a solution of this equation}
OR
x^2 + 10x - 39 = (3)^2 + 10(3) - 39 = 9 + 30 - 39 = 0 {therefore 3 is a solution of this equation}
METHOD TWO: QUADRATIC FORMULA The Quadratic Formula is very easy to use once you know it's structure:
x = (-b +/- _/b^2 - 4ac) / 2a
From here, all you have to do is substitute the variables:
Now, the +/- (plus or minus) symbol diverts your options into two. We can find one value of x by adding -10 and 16, and another value of x by subtracting 16 from -10:
x = (-10 + 16) / 2 = 6 / 2 = 3 {which we know is a solution of the equation}
OR
x = (-10 - 16) / 2 = (-26) / 2 = -13 {which we know is another solution of the equation}
Therefore, the solutions to b (in terms of your equation) are - 13 and 3.
