AMC 8 2019 Problem 20 | Fundamental Theorem of Algebra

Try out this beautiful algebra problem number 2 from AMC 8 2019 based on the Fundamental Theorem of Algebra.

AMC 8 2019 Problem 20:

How many different real numbers $x$ satisfy the equation $(x^{2}-5)^{2}=16?$

$\textbf{(A) }0$
$\textbf{(B) }1$
$\textbf{(C) }2$
$\textbf{(D) }4$
$\textbf{(E) }8$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is (D) 4

AMC 8, 2019, Problem 20

Try with Hints

The given equation is

$(x^2-5)^2 = 16$

and that means

$x^2-5 = \pm 4$

Among both cases, if

$x^2-5 = 4$

then,

$x^2 = 9 \implies x = \pm 3$

and that means we have 2 different real numbers that satisfy the equation.

and if we take another case, then

$x^2-5 = -4$

and so,

$x^2 = 1 \implies x = \pm 1$

and that means we have 2 different real numbers in this `case too that satisfy the equation. So total 2+2=4 real numbers that satisfy the equation.

AMC 8 2019 Problem 16 | Algebra Problem

Try this beautiful Number Theory problem from the AMC 2019 Problem 16. You may use sequential hints to solve the problem.

Algebra Question - AMC 8, 2019 Problem 16

Qiang drives 15 miles at an average speed of 30 miles per hour. How many additional miles will he have to drive at 55 miles per hour to average 50 miles per hour for the entire trip?
(A) 45
(B) 62
(C) 90
(D) 110
(E) 135

Key Concepts

Algebra

Value

Average Speed

Check the Answer

Answer: is (D) 110

AMC 8, 2019, Problem 16

Try with Hints

Among the options, there is only one option which is divisible by 55 and that is 110.

That option tells the travel hour is 2.

Qiang drives 15 miles at an average speed of 30 miles per hour.

And we know, Average speed = Total Distance/Total Time

So, by the formula,

$\frac{15}{30} + \frac{110}{55} = \frac{5}{2}$

In this case, If we consider the whole journey, Total Distance is (110+15)=125

And as Qiang has to drive at 50 miles per hour for the entire trip, and as Average speed = Total Distance/Total Time ,

$\frac{125}{50} = \frac{5}{2}$

As both are same , our answer 110 is established.

AMC 8 2019 Problem 17 | Value of Product

Try out this beautiful algebra problem from AMC 8, 2019 based on finding the value of the product. You may use sequential hints to solve the problem.

AMC 8 2019: Problem 17

What is the value of the product

$\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right) ?$

(A) $\frac{1}{2}$

(B) $\frac{50}{99}$

(D) $\frac{100}{99}$

(E) $50$

Key Concepts

Algebra

Value

Telescoping

Check the Answer

Answer: is $\frac{50}{99}$

AMC 8, 2019, Problem 17

Try with Hints

We write

$\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$

in a different form like

$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$

All of the middle terms eliminate each other, and only the first and last term remains i.e.

$\frac{1}{2} \cdot \frac{100}{99}$

$\frac{1}{2} \cdot \frac{100}{99}=\frac{50}{99}$

and that is the final answer.

AMC 8 2019 Problem 20 | Fundamental Theorem of Algebra