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 $x \neq 1$, solve for $x$ in $2 \sqrt{x} + \frac{3}{\sqrt{x}} = 5$.

Solution
Multiplying both sides by $\sqrt{x}$, we obtain

$2x + 3 = 5\sqrt{x}$

Squaring both sides of the equation, we get

$4x^2 + 12x + 9 = 25x$

$4x^2 - 13x + 9 = 0$.

Factoring, we have

$(4x - 9)(x - 1) = 0$

So, $x = \frac{9}{4}$ or $x = 1$.

But from the given above, $x \neq 1$, so the only solution is $x = \frac{9}{4}$

12.) Evaluate $\sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$.

Solution

Let $x = \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$

$x^2 = 2 + \sqrt{2 + \sqrt{2 + 2 \sqrt{2 + \cdots }}}$

$x^2 = 2 + x$

$x^2 - x - 2 = 0$

$(x - 2)(x + 1) = 0$

$x = 2$ or $x = -1$

Answer: $x = 2$ or $x = -1$

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

Solution

Let $x$ is equal to the smaller number and $x + 1$ be the larger number.

$x(x + 1) = x^2 + x = 506$

$x^2 + x - 506 = 0$

$(x - 22)(x - 23) = 0$

$x = 22$ or $x = 23$.

14.) If $c > a > 0$ and if $a - b + c = 0$, find the larger root of $ax^2 + bx + c = 0$.

Solution

From the given, $a - b + c = 0$, so $b = a + c$.

$x = \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

$= \dfrac{-(a + c) \pm \sqrt{(a + c)^2 - 4ac}}{2a}$

$= \dfrac{-a - c \pm \sqrt{a^2 - 2ac + c^2} }{2a}$

$= \dfrac{(-a - c) \pm a - c}{2a}$.

Simplifying, we get the two roots $x_1$ and $x_2$.

$x_1 = \dfrac{-2c}{2a} = -\dfrac{c}{a}$

$x_2 = \dfrac{-2a - 2c}{2a} = \dfrac{-a - c}{a}$

Since $c > a > 0$, the larger root is $-\dfrac{c}{a}$.

Answer: $-\dfrac{c}{a}$

15.) Solve for $x$ in $2x^2 + x < 6$.

Solution

Subtracting 6 from both sides,

$2x^2 + x - 6 < 0$

$(2x - 3)(x + 2) < 0$

$-2 < x < \frac{3}{2}$

Answer: $-2 < x < \frac{3}{2}$

16.) Solve for real numbers $x$ satisfying the inequality $x - 2\sqrt{x} \leq 3$.

Solution

$x - 3 \leq 2 \sqrt(x)$

Squaring both sides of the inequality,

$x^2 - 6x + 9 \leq 4x$

$x^2 - 10x + 9\leq 0$

$(x - 1)(x - 9)\leq 0$

This can be split to (1) $(x - 1)(x - 9) = 0$ and $(x - 1)(x - 9)< 0$. From (1), we already know that $x = 1$ and $x = 9$ are solutions. For the inequality, we can check the intervals $(- \in, 1)$, $(1, 9)$ and $(9, \infty)$. 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:  $1 \leq x \leq 9$ or $[1,9]$ in interval notation.

17.) Find the minimum value of x^2 – 8x + 3.

Solution

Let $y = x^2 - 8x + 3$.

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).

$x^2 - 8x + 3 = (x - 4)^2 + 3 - 16 = (x - 4)^2 - 13$

So, h = 4 and k = – 13. The minimum value is -13.

18.) Find the smallest value of $x + \dfrac{5}{x}$ for all real number $x$.

19.) Solve for $b$ in the equation $(x + 1)(x + a) = x^2 + bx + 3$.

Solution

$x^2 + x + ax + a = x^2 + bx + c$

$x + ax + a = bx + 3$.

Factoring the left side, we have

$x (1 + a)x + a = bx + 3$, therefore, a = 3 and b = 1 + 3 = 4.

20.) Write the quadratic equations with integer coefficients whose roots are reciprocal of the roots of $2x^2 - 3x + 1 = 0$.

Solution

$2x^2 - 3x + 1 = (2x - 1)(x - 1)= 0$

$x = \frac{1}{2}$ or $x = 1$.

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

$(x - 1)(x - 2) = x^2 - 3x + 2 = 0$

Answer: $x^2 - 3x + 2 = 0$.

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5 thoughts on “Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 2”

1. 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.

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