# 2015 MTAP Reviewer for Grade 9 With Solutions(11-15)

We’re done with 20% of the elimination paper of 2015 MMC (MTAP)  for grade 9, now we will continue learning how to deal with mind boggling problems to win the competition next year.

Problem 11:

If $x\ne 0$, solve for x in $2\sqrt{x}+\dfrac{3}{\sqrt{x}}=5$

Solution:

To solve equations, we always eliminate fractions. By multiplying the whole equation by $\sqrt{x}$ we can remove all fractions.

$(2\sqrt{x}+\dfrac{3}{\sqrt{x}})\sqrt{x}=5\sqrt{x}$

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

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

Now, since the right side of equation is radical, we raise both sides by 2,

$(2x+3)^2=(5\sqrt{x})^2$

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

$4x^2+12x-25x+9=0$

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

By factoring we have,

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

These factors will immediately tell us that the values of x are 1 and 9/4. By quick check however, 9/4 fails to satisfy the original equation. Thus, the only solution is 1.

Problem 12:

Evaluate  $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$

Solution:

This is a nested radical and I already have a tutorial on how to solve this problem. If you’re used to see problem like this, you might be able to solve it in less than 3 seconds. For real.

To solve this though, we let x be equal to $\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$

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

Square both sides,

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

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

Now, recall that $x=\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}$

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

$x^2=2+(x)$

$x^2=2+x$

$x^2-x-2=0$

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

$x=2,-1$

The square root of a positive number will always be positive. Thus, the answer  is 2. -1 is an extraneous root.

Problem 13:

Find the two consecutive integers whose product is 506.

Solution:

Let x be the smaller integer. Since we are looking for the two CONSECUTIVE integers, the other number must be x+1.

It is stated that their product is 506, thus

$x(x+1)=506$

$x^2+x-506=0$

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

$x=22,-23$

Therefore, the numbers that we are looking for are are 22 and 23.

Problem 14:

If $c>a>0$ and if $a-b+c=0$, what is the larger root of  $ax^2+bx+c=0$

Solution:

Let x and y be the roots of equation.

By Vieta’s formula, we have

$xy=\dfrac{c}{a}$ *

$x+y=\dfrac{-b}{a}$ **

From the relationship of the coefficients, we solve for c in terms of a and b.

$a-b+c=0$

$c=b-a$***

We substitute *** to c of *,

$xy=\dfrac{b-a}{a}$

$xy=\dfrac{b}{a}-\dfrac{a}{a}$

$xy=\dfrac{b}{a}-1$ ****

From ** we solve for b/a,

$x+y=\dfrac{-b}{a}$

$-x-y=\dfrac{b}{a}$

We substitute this to ****

$xy=\dfrac{b}{a}-1$

$xy=-x-y-1$

$xy+x+y+1=0$

By factoring,

$x(y+1)+(y+1)=0$

$(y+1)(x+1)=0$

$x=-1,y=-1$

Since the roots are equal, there is no greater root. The answer is -1.

Problem 15:

Solve for x in $2x^2+x<6$

Solution:

We also provided the easiest way to solve this quadratic inequality in this site. But let us demonstrate the answer to this problem.

$2x^2+x<6$

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

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

In this format, the answer is in the form of a<x<b, where b>a and a and b are the roots of inequality.

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

The roots are supposed to be 3/2 and -2. Thus, the solution set is -2<x<3/2 or (-2,3/2)

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.