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3 years ago

3(12x2+x+1)+12(12x2+x+1)=15(12x2+x+1)

Mathematics

2 answers:

Morgarella [4.7K]3 years ago

8 0

Answer:

infinite solutions

Step-by-step explanation:

3(12x2+x+1)+12(12x2+x+1)=15(12x2+x+1)

Combine like terms

3(12x2+x+1)+12(12x2+x+1)=15(12x2+x+1)

15(12x2+x+1) = 15(12x2+x+1)

Divide by 15

(12x2+x+1) = (12x2+x+1)

Since they are the same

x can be any real value

Vsevolod [243]3 years ago

6 0

Answer:

All real values of x

Step-by-step explanation:

3(12x²+x+1)+12(12x²+x+1)=15(12x²+x+1)

(12x² + x + 1)(3+12) = 15(12x²+x+1)

15(12x²+x+1) = 15(12x²+x+1)

Since both sides are identical, all real values of x will satisfy the equation

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Vlad [161]

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Step-by-step explanation:

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Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

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These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

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Evaluating $y(0)=2$ :

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$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

3 years ago