Find one value of x that is a solution to the equation: {:[(x^(2)-7)^(2)+2x^(2)-14=0],[x=◻]:}

Expand and combine terms: We are given the equation [(x^2 - 7)^2 + 2x^2 - 14 = 0]. Let's first expand the squared term and then combine like terms. [(x^2 - 7)^2] becomes [x^4 - 14x^2 + 49]. So the equation becomes [x^4 - 14x^2 + 49 + 2x^2 - 14 = 0].Simplify the equation: Combine the x^2 terms and the constant terms. [x^4 - 14x^2 + 2x^2] becomes [x^4 - 12x^2]. [49 - 14] becomes [35]. So the equation now is [x^4 - 12x^2 + 35 = 0].Substitute and factor: This is a quadratic equation in terms of x^2. Let's substitute y for x^2, so we have [y^2 - 12y + 35 = 0]. Now we need to factor this quadratic equation.Find possible solutions for y: We are looking for two numbers that multiply to 35 and add up to -12. These numbers are -5 and -7. So we can write the factored form as [(y - 5)(y - 7) = 0].Solve for x in each case: Now we have two possible solutions for y: y = 5 or y = 7. Remember that y was a substitute for x^2, so we now have [x^2 = 5] or [x^2 = 7].Final solutions: We will solve for x in each case. Starting with [x^2 = 5], we take the square root of both sides to get [x = √5] or [x = -√5].Now we solve [x^2 = 7] by taking the square root of both sides to get [x = √7] or [x = -√7].We have found four possible values for x: [√5, -√5, √7, -√7]. We can choose any one of these as a correct solution to the original equation.

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Algebra 1

Solve a quadratic equation by factoring

Full solution

Q. Find one value of

x

that is a solution to the equation:

\newline

\left(x^{2}-7\right)^{2}+2x^{2}-14=0

\newline

x=\square

Expand and combine terms: We are given the equation $[(x^2 - 7)^2 + 2x^2 - 14 = 0]$ . Let's first expand the squared term and then combine like terms. $[(x^2 - 7)^2]$ becomes $x^4 - 14x^2 + 49$ . So the equation becomes $x^4 - 14x^2 + 49 + 2x^2 - 14 = 0$ .
Simplify the equation: Combine the $x^2$ terms and the constant terms. $\newline$ $[x^4 - 14x^2 + 2x^2]$ becomes $[x^4 - 12x^2]$ . $\newline$ $[49 - 14]$ becomes $[35]$ . $\newline$ So the equation now is $[x^4 - 12x^2 + 35 = 0]$ .
Substitute and factor: This is a quadratic equation in terms of $x^2$ . Let's substitute $y$ for $x^2$ , so we have $[y^2 - 12y + 35 = 0]$ . $\newline$ Now we need to factor this quadratic equation.
Find possible solutions for $y$ : We are looking for two numbers that multiply to $35$ and add up to $-12$ . These numbers are $-5$ and $-7$ . $\newline$ So we can write the factored form as $[(y -$ $5$ )(y - $7$ ) = $0$ ].
Solve for x in each case: Now we have two possible solutions for y: $y = 5$ or $y = 7$ . $\newline$ Remember that y was a substitute for $x^2$ , so we now have $x^2 = 5$ or $x^2 = 7$ .
Final solutions: We will solve for $x$ in each case. Starting with $x^2 = 5$ , we take the square root of both sides to get $x = \sqrt{5}$ or $x = -\sqrt{5}$ .
Final solutions: We will solve for x in each case. Starting with $x^2 = 5$ , we take the square root of both sides to get $x = \sqrt{5}$ or $x = -\sqrt{5}$ .Now we solve $x^2 = 7$ by taking the square root of both sides to get $x = \sqrt{7}$ or $x = -\sqrt{7}$ .
Final solutions: We will solve for x in each case. Starting with $x^2 = 5$ , we take the square root of both sides to get $x = \sqrt{5}$ or $x = -\sqrt{5}$ . Now we solve $x^2 = 7$ by taking the square root of both sides to get $x = \sqrt{7}$ or $x = -\sqrt{7}$ . We have found four possible values for x: $\sqrt{5}$ , $-\sqrt{5}$ , $\sqrt{7}$ , $-\sqrt{7}$ . We can choose any one of these as a correct solution to the original equation.