501 Questions To Master GED Test Mathematics_pag197

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501 Questions to Master the GED® Mathematical Reasoning Test
 248. The correct answer is x = –5. This question requires you to solve 
x2 – 15x = 100 and identify the larger solution. First rearrange 
x2 – 15x = 100 so that a is positive and all the terms are on one 
side of the equation:
 x2 – 15x = 100
 –100 –100
 x2 – 15x – 100 = 0
 The factored form of this will be (x + m)(x + n) = 0, where mn = 
c and m + n = b. Identify that c = –100 and b = –15. Since the two 
factors must multiply to a negative and add to a negative, we can 
identify that one factor must be negative and the other must be 
positive. Create a list of factor pairs that multiply to –100 with the 
larger factor being negative (that way they will add to a negative b): 
{1,–100}, {2,–50}, {4,–25}, {5,–20}, and {10,–10}. The factors 5 and 
–20 will multiply to –100 and add to –15, so substitute 5 and –20 
into (x + m)(x + n): (x + 5)(x – 20) = 0. Now solve for (x + 5) = 0 and 
(x – 20) = 0 to arrive at the two solutions of x = –5 and x = 20. Since 
the question asked for the smallest solution, the correct answer will 
be x = –5.
 249. The correct answer is choice b. This is a quadratic equation and 
can be solved in several ways: factoring, completing the square, or 
using the quadratic formula. However, looking at the terms –5x 
and –6, it is apparent that factors of 6 can add to 5, so we will factor 
this quadratic to get the solutions. First, we set the equation equal 
to 0: x2 – 5x + 6 = 0. We want to find the two values of x where the 
equation is equal to 0. By factoring, we get (x – 2)(x – 3). Multiply 
it out if necessary to check that it in fact yields the original equa-
tion. Anything multiplied by 0 is 0, so for what values of x would 
we get an answer of 0? When x = 2 and when x = 3. 
 Choices a, c, and d have incorrectly placed positive and negative 
signs.
 250. The correct answer is 3 or 4. Since you are not permitted to 
repeat one of the existing coordinate pairs, you cannot use 6, 5, 2, 
or 1 as x-values since those are the existing x-values in unique pair-
ings. Only the coordinates 3 or 4 could be chosen for the x value, 
and then any of the other terms could work for the corresponding 
y-values.
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