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Eva8 [605]
3 years ago
5

Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

Mathematics
1 answer:
BigorU [14]3 years ago
4 0

Answer:

x=2  true solution

x=-3 extraneous

Step-by-step explanation:

sqrt(x+7) -1 = x

Add 1 to each side

sqrt(x+7) -1+1 = x+1

sqrt(x+7)  = x+1

Square each side

(sqrt(x+7))^2  = (x+1)^2

x+7 = x^2 +2x+1

Subtract x from each side

7 = x^2 +x +1

Subtract 7 from each side

0 = x^2 +x - 6

Factor

0 = (x+3)(x-2)

Using the zero product property

x+3 = 0   x-2 =0

x=-3  x=2

Check solutions

x=-3

sqrt(-3+7) -1 = -3

sqrt(4) -1 = -3

3 =-3  extraneous

x=2

sqrt(2+7) -1 = 2

sqrt(9) -1 = 2

3 -1 =2

2 =2  true

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Please help me. I am so stuck.<br>​
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Converges to -25.

Step-by-step explanation:

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6 0
3 years ago
Use the variation of parameters method to solve the DR y" + y' - 2y = 1
postnew [5]

Answer:

y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Step-by-step explanation:

As given in question, we have to find the solution of differential equation

y"+y'-2y=1

by using the variation in parameter method.

From the above equation, the characteristics equation can be given by

D^2+D-2\ =\ 0

=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}

=>\ D=\ -2\ or\ 1

Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by

y_c(t)\ =\ C_1e^{-2t}+C_2e^t

Let's assume that

     y_1(t)=e^{-2t}          y_2(t)=e^t

=>\ y'_1(t)=-2e^{-2t}        y'_2(t)=e^t

   and g(t)=1

Now, the Wronskian can be given by

W=y_1(t).y'_2(t)-y'_1(t).y_2(t)

   =e^{-2t}.e^t-e^t(-e^{-2t})

   =e^{-t}+2e^{-t}

   =3e^{-t}

Now, the particular solution can be given by

y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}

=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}

=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}

=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}

=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

Now, the complete solution of the given differential equation can be given by

y(t)\ =\ y_c(t)+y_p(t)

      =C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}

5 0
3 years ago
Solve for x.
Flauer [41]
I have to say its
<span>−64 ​ and 64</span>
4 0
3 years ago
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