This is the second part (questions 11-20) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. The first part can be read here.
Although reasonable care has been given to make the solution accurate as possible, the solver is also human. Please comment below if you see any errors.
11.) If , solve for in .
Multiplying both sides by , we obtain
Squaring both sides of the equation, we get
Factoring, we have
So, or .
But from the given above, , so the only solution is
12.) Evaluate .
13.) Find two positive consecutive integers whose product is 506.
Let is equal to the smaller number and be the larger number.
Answer: 22 and 23
14.) If and if , find the larger root of .
From the given, , so .
Simplifying, we get the two roots and .
Since , the larger root is .
15.) Solve for in .
Subtracting 6 from both sides,
16.) Solve for real numbers satisfying the inequality .
Squaring both sides of the inequality,
This can be split to (1) and . From (1), we already know that $x = 1$ and are solutions. For the inequality, we can check the intervals , and . Upon checking, it can be seen that only (1,9) satisfy the inequality. Therefore, the interval that satisfies the inequality above is [1,9].
Answer: or in interval notation.
17.) Find the minimum value of x^2 – 8x + 3.
The graph of the function opens upward, so we transform it to the vertex form a(x – h)^2 + k = 0 to get the minimum value which is (h,k).
So, h = 4 and k = – 13. The minimum value is -13.
18.) Find the smallest value of for all real number .
19.) Solve for in the equation .
Factoring the left side, we have
, therefore, a = 3 and b = 1 + 3 = 4.
20.) Write the quadratic equations with integer coefficients whose roots are reciprocal of the roots of .
So, we create a quadratic equation whose roots are 2 and 1. That is
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