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2 years ago

Find one value of xxx that is a solution to the equation: (5x+2)^2+15x+6=0(5x+2) 2 +15x+6=0

1 answer:

5 0

An equation is formed of two equal expressions. The solution of the equation (5x+2)² + 15x + 6=0 is -0.4 and -1.

<h3>What is an equation?</h3>

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

The solution to the equation,

(5x+2)² + 15x + 6=0

25x² + 4 + 20x + 15x + 6 = 0

25x² + 35x + 10 = 0

Plotting the equation on the graph, the value of x is,

x = -0.4, -1

Hence, the solution of the equation (5x+2)² + 15x + 6=0 is -0.4 and -1.

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Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

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Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

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The pattern of the normalized differential equation can be represented as:

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This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

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