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)

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