# 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.