# 2015 MTAP Reviewer for Grade 9 Solutions(16-20)

This is the 4th tutorial series about the solution to 2015 MTAP(MMC) elimination paper for grade 9. Links to other pages included here where you can enhance you knowledge in specific subject matter.

Problem 16:

Solve for real numbers satisfying the inequality  $x-2\sqrt{x}\le 3$

Solution:

By rearranging,

$x-2\sqrt{x}\le 3$

$x-2\sqrt{x}-3\le 0$

$(\sqrt{x}-3)(\sqrt{x}+2)\le 0$

Solving for x:

$\sqrt{x}-3\le 0,\sqrt{x}+2\ge 0$

$\sqrt{x}\le 3,\sqrt{x}\ge -2$

Obviously, -2 is an extraneous root since we have square root of x in the left side of equation and the square root of any number will always be greater than 0. Thus,

$\sqrt{x}\le 3$

$(\sqrt{x})^2\le 3^2$

$\boxed{x\le 9}$

Check this tutorial on how to deal with quadratic inequality flawlessly.

Problem 17:

Find the minimum value of   $x^2-8x+3=0$

Solution:

Minimum value of quadratic equation can be found using the following formula,

$\min=\dfrac{4ac-b^2}{4a}$

where a,b, and c are the coefficients of quadratic equation.

By substitution,

$\min=\dfrac{4(1)(3)-(8)^2}{4(1)}$

$\min=\dfrac{12-64}{4}$

$\min=3-16$

$\boxed{\min=-13}$

Problem 18:

Find the smallest value of  $x+\dfrac{5}{x}$

Solution:

By AM-GM Inequality theorem we have,

$\dfrac{x_1+x_2+x_3+\cdots+x_n}{n}\ge \sqrt[n]{x_1\cdot x_2\cdot x_3\cdots}$

Applying the theorem,

$\dfrac{x+\dfrac{5}{x}}{2}\ge\sqrt{x\cdot \dfrac{5}{x}}$

$\boxed{x+\dfrac{5}{x}\ge 2\sqrt{5}}$

Thus, the minimum value of x+5/x is $\boxed{2\sqrt{5}}$

Problem 19:

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

Solution:

$(x+1)(x+a)=x^2+bx+3$

$x^2+x+ax+a=x^2+bx+3$

$x^2+(a+1)x+a=x^2+bx+3$

By comparison, we can say that b=a+1. But a=3(constant part), hence

$b=a+1$

$b=3+1$

$\boxed{b=4}$

Problem 20:

Write the quadratic equation with integer coefficients whose roots are the reciprocals of the roots of $2x^2-3x+1=0$

Solution:

This problem actually can be rigidly solved using Vieta’s formula but there is a hippest way to solve problem like this where we can solve it in just a matter of three to 4 lines long.

Let y be the roots of the desired equation. Since the given equation has  a root of x, we can establish the following relation based on the condition given(reciprocal of the roots).

$y=\dfrac{1}{x}$

$x=\dfrac{1}{y}$

Now, we substitute this to the given equation,

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

$2(\dfrac{1}{y})^2-3(\dfrac{1}{y})+1=0$

$\dfrac{2}{y^2}-\dfrac{3}{y}+1=0$

Simplifying to have an integer coefficients we have,

$2-3y+y^2=0$

Dropping y back to x,

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

$\boxed{x^2-3x+2=0}$