AWhich of the possible methods seems easiest for you to use?
a
B
b Analyze the form of the equation and discard the methods that are not applicable.
A
a See solution.
B
b See solution.
Practice makes perfect
a Let's start by taking a look at the equation given.
k^2-3k=304Note that we can rewrite the quadratic equation in standard form obtaining a trinomial of the form k^2+ bk+ c on the left-hand side.
k^2-3k=304
⇕
k^2+( -3)k+( -304) = 0
We can always choose the method that seems easier for us and preferences may vary, however, this is an expression that can be factored. Therefore, it is convenient to solve the equation by factoring. Since -19+16= -3, and - 19 * 16 = - 304, we can use them to factor our equation.
k^2-3k - 304 = 0
⇕
(k-19)(k+16) = 0
By the Zero Product Property we know that the solutions satisfy k-19 = 0 and k+16=0. Therefore, the solutions are k=19 and k=- 16.
b Let's start by taking a look at the equation given.
t^2-6t+16=0
Since this equation is not of the form x^2=k, we cannot simply solve it by just using square roots. Furthermore, since there are no factors of 16 that add up to -6, it cannot be factored as a product of the form shown below, with p and q integers.
(t-p)(t-q)Therefore, we have to solve it by completing the square. To do this, we first need to isolate the terms containing the variable t.
t^2-6t=- 16
Now, we have an expression of the form at^2+bt on the left-hand side.
at^2+bt
1t^2+(- 6)t
As we can see, in this case a=1 and b=- 6. Now, we can add ( b2)^2= ( - 62)^2 to both sides to complete the square. Let's simplify this quantity first.
The next step would be to calculate the square root of both sides. However, this is not possible since the right-hand is negative. Therefore, we can know that the given equation has no real solutions.
