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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


1st Hint

The given equation is

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

and that means

$x^2-5 = \pm 4$

2nd Hint

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.

Final Step

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.

Cheenta Numerates Program for AMC - AIME

Subscribe to Cheenta at Youtube


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


1st Hint

The given equation is

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

and that means

$x^2-5 = \pm 4$

2nd Hint

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.

Final Step

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.

Cheenta Numerates Program for AMC - AIME

Subscribe to Cheenta at Youtube


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