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3 years ago

Solve by factoring. p⁴=-25p²

Mathematics

1 answer:

3241004551 [841]3 years ago

5 0

Answer:

p=5i

Step-by-step explanation:

(p^(4)=-25p^(2))/p^(2)

p^(2)=-25

Square root both sides

p=5i

i is the square root of -1, so 5i is five times that much.

The square root of a negative number in not real. That is why it is i, and not a number.

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Jayden and Sheridan both tried to find the missing side of the right triangle. A right triangle is shown. One leg is labeled as

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Answer:

Sheridan's Work is correct

Step-by-step explanation:

we know that

The lengths side of a right triangle must satisfy the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

where

a and b are the legs

c is the hypotenuse (the greater side)

In this problem

Let

$a=7\ cm\\c=13\ cm$

substitute

$13^{2}=7^{2}+b^{2}$

Solve for b

$169=49+b^{2}$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

we have that

<em>Jayden's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\b=13\ cm$

substitute and solve for c

$7^{2}+13^{2}=c^{2}$

$49+169=c^{2}$

$218=c^{2}$

$c=\sqrt{218}\ cm$

$c=14.76\ cm$

Jayden's Work is incorrect, because the missing side is not the hypotenuse of the right triangle

<em>Sheridan's Work</em>

$a^{2}+b^{2}=c^{2}$

$a=7\ cm\\c=13\ cm$

substitute

$7^{2}+b^{2}=13^{2}$

Solve for b

$49+b^{2}=169$

$b^{2}=169-49$

$b^{2}=120$

$b=\sqrt{120}\ cm$

$b=10.95\ cm$

therefore

Sheridan's Work is correct

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Solve by factoring. p⁴=-25p²​

Solve by factoring. p⁴=-25p²