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Let
p=x^(2)-7.
Which equation is equivalent to
(x^(2)-7)^(2)-4x^(2)+28=5 in terms of
p ?
Choose 1 answer:

p^(2)-4p+23=0

p^(2)+4p-5=0

p^(2)-4p-5=0

p^(2)+4p+23=0

Let $p=x^{2}-7$ . Which equation is equivalent to $(x^{2}-7)^{2}-4x^{2}+28=5$ in terms of $p$ ? Choose $1$ answer: $\newline$ $p^{2}-4p+23=0$ $\newline$ $p^{2}+4p-5=0$ $\newline$ $p^{2}-4p-5=0$ $\newline$ $p^{2}+4p+23=0$

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Find derivatives of using multiple formulae

Full solution

Q. Let

p=x^{2}-7

. Which equation is equivalent to

(x^{2}-7)^{2}-4x^{2}+28=5

in terms of

p

? Choose

1

answer:

\newline

p^{2}-4p+23=0

\newline

p^{2}+4p-5=0

\newline

p^{2}-4p-5=0

\newline

p^{2}+4p+23=0

Substitute $p$ into equation: Given $p = x^2 - 7$ , we want to express the equation $(x^2 - 7)^2 - 4x^2 + 28 = 5$ in terms of $p$ .
Express $-4x^2$ in terms of $p$ : First, we substitute $p$ into the equation where $x^2 - 7$ appears: $\newline$ $(p)^2 - 4x^2 + 28 = 5$
Distribute $-4$ across terms: Now, we need to express $-4x^2$ in terms of $p$ . Since $p = x^2 - 7$ , we can rewrite $x^2$ as $p + 7$ : $(p)^2 - 4(p + 7) + 28 = 5$
Simplify by canceling terms: Next, we distribute the $-4$ across the terms in the parentheses: $(p)^2 - 4p - 28 + 28 = 5$
Set equation to zero: The $-28$ and $+28$ cancel each other out, simplifying the equation to: $(p)^2 - 4p = 5$
Final equivalent equation: To set the equation to zero, we subtract $5$ from both sides: $\newline$ $(p)^2 - 4p - 5 = 0$
Final equivalent equation: To set the equation to zero, we subtract $5$ from both sides: $\newline$ $(p)^2 - 4p - 5 = 0$ We have now expressed the original equation in terms of $p$ , and the equivalent equation is: $\newline$ $p^2 - 4p - 5 = 0$

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