# Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2

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 .

**Solution**

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

**Answer: 9/4 **

**12.)** Evaluate .

**Solution**

Let

or

**Answer:** or

**13.)** Find two positive consecutive integers whose product is 506.

**Solution**

Let is equal to the smaller number and be the larger number.

or .

**Answer**: 22 and 23

**14.)** If and if , find the larger root of .

**Solution**

From the given, , so .

.

Simplifying, we get the two roots and .

Since , the larger root is .

**Answer:**

**15.)** Solve for in .

**Solution**

Subtracting 6 from both sides,

**Answer:**

16.) Solve for real numbers satisfying the inequality .

**Solution**

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.

**Solution**

Let .

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.

**Answer**: -13

**18.)** Find the smallest value of for all real number .

**Answer**: 4.47

**19.)** Solve for in the equation .

**Solution**

.

Factoring the left side, we have

, therefore, a = 3 and b = 1 + 3 = 4.

**Answer:** 4

**20.)** Write the quadratic equations with integer coefficients whose roots are reciprocal of the roots of .

**Solution**

or .

So, we create a quadratic equation whose roots are 2 and 1. That is

**Answer**: .

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I was wondering in question 16, why 0 is not included in solution set infact 0-2sqrt(0)<=3. I check that in maple, mathlab and symbolab, 0 is included.

Question 14, I think the larger root is x = -1. Kindly check your second root. That is plus minus ( a-c ).

correction – ( a – c )

Ang nakuha kong answer sa number 19, b = 4

(x+1)(x+a) = x^2 + bx + 3

x^2 + x + ax + a = x^2 + bx + 3

x^2 +(1+a)x + a = x^2 + bx + 3

Meaning, (1+a) = b and a = 3

(1+3) = b

4 = b