How do I solve Quadratic equations by factoring?
(for example how would I solve n2 - 10n = -21 if I was supposed to get the answers 7 and 3?
first, set everything equal to zero.
n^2 - 10n + 21 = 0
now, since (n-a)(n-b) = n^2 - (a+b)n + ab, we are looking for two numbers a and b, which multiply to 21 and add to 10.
Now, the only factor of 21 are 7 and 3. Luckily, they add to 10, so we have
(n-7)(n-3) = n^2 - 10n + 21
Now, back to step 1. Why set everything to zero? We now have
(n-7)(n-3) = 0
the only way two numbers can multiply to zero is if one or the other of them is zero. So, we have either
n-7 = 0 ==> n=7
or
n-3 = 0 ==> n=3
Those are the solutions to
(n-7)(n-3) = 0, which is just a rewriting of the original equation.
To solve quadratic equations by factoring, follow these steps:
Step 1: Rewrite the equation in standard form. In your example, the equation is already in standard form: n^2 - 10n = -21.
Step 2: Move all terms to one side of the equation, so that it equals zero. In this case, add 21 to both sides to get: n^2 - 10n + 21 = 0.
Step 3: Factor the quadratic expression completely. Look for two numbers whose product is equal to the constant term (21 in this case) and whose sum is equal to the coefficient of the linear term (-10n in this case).
The factors of 21 are: 1, 3, 7, and 21.
Testing different factor combinations, we find that 3 and 7 satisfy both conditions (3 * 7 = 21 and 3 + 7 = 10).
So, the factored form of the quadratic equation is: (n - 3)(n - 7) = 0.
Step 4: Set each factor equal to zero and solve for n. In this case, we have:
n - 3 = 0 or n - 7 = 0
Solving each equation gives us:
n = 3 or n = 7
Therefore, the solutions to the quadratic equation n^2 - 10n = -21 are n = 3 and n = 7.