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: (A) p^(2)-4p+23=0 (B) p^(2)+4p-5=0 (C) p^(2)-4p-5=0 (D) p^(2)+4p+23=0

Substitute p into equation: Given p = x^2 - 7, we need to express the equation (x^2 - 7)^2 - 4x^2 + 28 = 5 in terms of p.Express -4x^2 in terms of p: First, let's substitute p into the equation where we see x^2 - 7. This gives us 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. Therefore, -4x^2 becomes -4(p + 7).Simplify the equation: Substitute -4(p + 7) into the equation in place of -4x^2. This gives us p^2 - 4(p + 7) + 28 = 5.Subtract 5 from both sides: Now, let's distribute the -4 across the terms inside the parentheses: p^2 - 4p - 28 + 28 = 5.Express equation in standard form: The -28 and +28 cancel each other out, simplifying the equation to p^2 - 4p = 5.To get the equation in standard form, we subtract 5 from both sides, resulting in p^2 - 4p - 5 = 0.We have now expressed the original equation in terms of p, and the equivalent equation is p^2 - 4p - 5 = 0, which corresponds to option (C).

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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:
(A)
p^(2)-4p+23=0
(B)
p^(2)+4p-5=0
(C)
p^(2)-4p-5=0
(D)
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: $\newline$ (A) $p^{2}-4p+23=0$ $\newline$ (B) $p^{2}+4p-5=0$ $\newline$ (C) $p^{2}-4p-5=0$ $\newline$ (D) $p^{2}+4p+23=0$

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Math Problems

Algebra 1

Solve a quadratic equation by factoring

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

(A)

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

\newline

(B)

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

\newline

(C)

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

\newline

(D)

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

Substitute $p$ into equation: Given $p = x^2 - 7$ , we need to express the equation $(x^2 - 7)^2 - 4x^2 + 28 = 5$ in terms of $p$ .
Express $-4x^2$ in terms of $p$ : First, let's substitute $p$ into the equation where we see $x^2 - 7$ . This gives us $p^2 - 4x^2 + 28 = 5$ .
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 $p + 7$ . Therefore, $-4x^2$ becomes $-4(p + 7)$ .
Simplify the equation: Substitute $-4(p + 7)$ into the equation in place of $-4x^2$ . This gives us $p^2 - 4(p + 7) + 28 = 5$ .
Subtract $5$ from both sides: Now, let's distribute the $-4$ across the terms inside the parentheses: $p^2 - 4p - 28 + 28 = 5$ .
Express equation in standard form: The $-28$ and $+28$ cancel each other out, simplifying the equation to $p^2 - 4p = 5$ .
Express equation in standard form: The $-28$ and $+28$ cancel each other out, simplifying the equation to $p^2 - 4p = 5$ .To get the equation in standard form, we subtract $5$ from both sides, resulting in $p^2 - 4p - 5 = 0$ .
Express equation in standard form: The $-28$ and $+28$ cancel each other out, simplifying the equation to $p^2 - 4p = 5$ .To get the equation in standard form, we subtract $5$ from both sides, resulting in $p^2 - 4p - 5 = 0$ .We have now expressed the original equation in terms of $p$ , and the equivalent equation is $p^2 - 4p - 5 = 0$ , which corresponds to option $(C)$ .