# 2015 MTAP Reviewer for Grade 9 Solutions(21-25)

This is already the 5th part of the reviewer solution series for grade 9 MMC. You may check the previous parts in this section.

Problem 21:

Compute the sum of all the roots of (x-2)(x+1)+(x-1)(x+4)=0

Solution:

To find the sum of the roots, we expand the equation,

$(x-2)(x+1)+(x-1)(x+4)=0$

$x^2-x-2+x^2+3x-4=0$

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

$x^2+x-3=0$

Now, the sum of the roots of this quadratic equation is just the negative ratio of b and a. Thus, the sum is -1.

Problem 22:

If r and s are the roots of $x^2+x-1=0$, evaluate $(r+s)^2$

Solution:

The sum of the roots of quadratic equation is again the negative ratio of b and a. Thus,

$r+s=-1$

This means that $\boxed{(r+s)^2=1}$.

Actually, this problem doesn’t really make sense at all. I was expecting a twist somewhere. They might mistyped this $(r+s)^2$ from this $(r-s)^2$. The latter is more challenging and adds flavor to the problem. Anyways, let’s proceed.

Problem 23:

For what value(s) of m are the roots of $(m-1)x^2-mx+1=0$ equal?

Solution:

For a quadratic equation to have an equal roots, the discriminant must be equal to 0. Thus,

$D=0$

$b^2-4ac=0$

$(-m)^2-4(m-1)(1)=0$

$m^2-4m+4=0$

$(m-2)(m-2)=0$

$\boxed{m=2}$

Problem 24:

It is known that y varies as the square of x and y=8 when x=1. What is y when x=8?

Solution:

Let k be the constant of proportionality. Since y varies as the square of x we have,

$y=kx^2$

Solving for k when  y=8 and x=1,

$8=k(1)^2$

$k=8$

Solving for y when x=8,

$y=kx^2$

$y=8(8)^2$

$y=8(64)$

$y=8(64)$

$\boxed{y=512}$

Problem 25:

Suppose that x and y are inversely proportional and are positive quantities. By what percent does y decrease if x is increased by 25%?

Solution:

We let k be the constant of proportionality and we establish the equation.

$y=\dfrac{k}{x}$

IF x in increased by 25%, the value of x will become 1.25x. Since the constant of proportionality is constant. We equate the ks of two situations.

$k_1=k_2$

$x_1y_1=y_2x_2$

$x_1y_1=y_2(1.25x_1)$

$y_2=\dfrac{x_1y_1}{1.25x_1}$

$y_2=\dfrac{y_1}{1.25}$

$y_2=0.8y_1$

From 1, the value of y decreases to 0.8. Thus, it decreased by 20%.

### 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.