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Sliva [168]
1 year ago
5

Solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.

Mathematics
1 answer:
tatyana61 [14]1 year ago
8 0

Answer: \large\boxed{\text{No Solution}}

Step-by-step explanation:

<u>Given equation</u>

\sqrt{t+1}+9=7

<u>Subtract 9 on both sides</u>

\sqrt{t+1}+9-9=7-9

\sqrt{t+1}=-2

<u>Conclusion</u>

\text{Since radical values are always positive, it is false that }\sqrt{t+1}\text{ will equal to -2}

\text{Therefore, it is }{\large\boxed{\text{No Solution}}}\text{ in the context of real numbers}

Hope this helps!! :)

Please let me know if you have any questions

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Answer:

P(x) = x^4 -16x^3 +76x^2 -72x -100

Step-by-step explanation:

The two roots 1-√3 and 1+√3 give rise to the quadratic factor ...

... (x -(1-√3))(x -(1+√3)) = (x-1)^2 -(√3)^2 = x^2 -2x -2

The complex root 7-i has a conjugate that is also a root. These two roots give rise to the quadratic factor ...

... (x -(7 -i))(x -(7 +i)) = (x-7)^2 -(i)^2 = x^2 -14x +50

The product of these two quadratic factors is ...

... P(x) = (x^2 -2x -2)(x^2 -14x +50) = x^4 +x^3(-14 -2) +x^2(50 +28 -2) +x(-100+28) -100

... P(x) = x^4 -16x^3 +76x^2 -72x -100

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3 years ago
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Step-by-step explanation:

The formula for the volume is given by

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Plugging in 216 for the volume, we end up with

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Step-by-step explanation:

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If y-3x=9, 7x+y=25, what is the value of x and y?
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\bold{\rm{Given}}\text{ : If y - 3x = 9, 7x + y = 25, what is the value of x and y}?

\bold{\rm{Solution}} \text{: We'll solve this by substitution method}

\dashrightarrow  y - 3x = 9 \\  \\  \dashrightarrow   \boxed{y = 9 + 3x } \qquad .. \sf eq(1)

\text{we got the value of y now getting the value of x}

\looparrowright 7x + y = 25 \\  \\  \looparrowright 7x = 25 - y \\  \\  \looparrowright  \boxed{x =   \frac{25 - y}{7}} \qquad .. \sf eq(2)

\text{Now, putting y = 9 + 3x in eq(2)}

\hookrightarrow  x =  \frac{25 - y}{7}   \\  \\ \hookrightarrow  x =  \frac{25 - 9 + 3x}{7}  \\  \\ \hookrightarrow  7x =  25 - 9 + 3x \\  \\ \hookrightarrow  7x - 3x =  16 \\  \\ \hookrightarrow  4x = 16 \\  \\ \hookrightarrow  x =  \frac{16}{4}  \\  \\ \star \quad  \boxed{ \green{ \frak{ x = 4}}}

\text{Now we know the value of x, for getting y we need to put x in eq(1)}

: \implies y = 9 + 3x \\  \\   : \implies y = 9 + 3(4) \\  \\  : \implies y = 9 + 12  \\  \\    \star \quad  \boxed{\frak{  \green{y = 21}}}

\text{Hence, values of x and y are} \:  \frak{ \green{4}}  \: \text{and} \:  \frak{ \green{21}} \: \text{respectively}.

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2 years ago
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