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10 votes
10 votes
2n²+3n=27

2t²-162=0
find the roots of each equation.​

User Ninju
by
2.8k points

1 Answer

17 votes
17 votes

Answer:


n = 3, n=-\frac92; t=9, t=-9

Explanation:

Second one is faster, so let's get rid of it first.

Divide by 2 to make numbers easier, and we get


t^2-81 = 0. You should recognize 81 is 9 squared, and both
-9 and
+9, when squared, give 81, we have our solution.

Now the first. I can't spot any quick trick to solve it, so quadratic formula it is. Let's rememer it: if


ax^2 +bx+c=0 then


x={{-b \pm √(b^2-4ac)}\over 2a

Let's bring the first equation in the standard form and calculate the quantity over the square root (usually called with the greek letter delta,
\Delta) to the side.


2n^2+3n-27 =0\\\Delta = (3)^2 -4(2)(-27) = 9+(8)(3)(9) = 9(1+24)= 9 \cdot25 =(3\cdot5)^2

now we can apply the formula:


n={{-3\pm15}\over4} Let's split the two cases now


n= \frac{-3+15}4 = \frac{12}4=3\\n= \frac{-3-15}4 = \frac{-18}4 =-\frac92

User Domnic
by
3.0k points