Find one value of x that is a solution to the equation: {:[(x^(2)+3)^(2)=4x^(2)+12],[x=◻]:}

Expand left side of equation: We are given the equation (x^2 + 3)^2 = 4x^2 + 12. To find the value of x, we need to simplify and solve the equation. First, let's expand the left side of the equation using the formula (a + b)^2 = a^2 + 2ab + b^2. (x^2 + 3)^2 = (x^2)^2 + 2(x^2)(3) + (3)^2Perform calculations: Now, let's perform the actual calculations: (x^2 + 3)^2 = x^4 + 2(3x^2) + 9 This simplifies to: x^4 + 6x^2 + 9Equate to right side of equation: Next, we equate the expanded form to the right side of the original equation: x^4 + 6x^2 + 9 = 4x^2 + 12Move terms to one side: Now, we need to move all terms to one side to set the equation to zero and solve for x: x^4 + 6x^2 + 9 - 4x^2 - 12 = 0Combine like terms: Simplify the equation by combining like terms: x^4 + 2x^2 - 3 = 0Substitute y = x^2: This is a quadratic equation in terms of x^2. Let's substitute y = x^2 to make it easier to solve: y^2 + 2y - 3 = 0Factor the quadratic equation: Now, we factor the quadratic equation: (y + 3)(y - 1) = 0Solve for y: Set each factor equal to zero and solve for y: y + 3 = 0 or y - 1 = 0Substitute back to find x: Solving for y gives us two solutions: y = -3 or y = 1Solve for x: Since y = x^2, we substitute back to find the values of x: x^2 = -3 (This is not possible since x^2 cannot be negative for real numbers.) x^2 = 1Solve for x when x^2 = 1: x = √1 or x = -√1 x = 1 or x = -1

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Math Problems

Algebra 1

Solve a quadratic equation by factoring

Full solution

Q. Find one value of

x

that is a solution to the equation:

\newline

\begin{array}{l} \left(x^{2}+3\right)^{2}=4 x^{2}+12 \\ x=\square \end{array}

Expand left side of equation: We are given the equation $(x^2 + 3)^2 = 4x^2 + 12$ . To find the value of $x$ , we need to simplify and solve the equation. $\newline$ First, let's expand the left side of the equation using the formula $(a + b)^2 = a^2 + 2ab + b^2$ . $\newline$ $(x^2 + 3)^2 = (x^2)^2 + 2(x^2)(3) + (3)^2$
Perform calculations: Now, let's perform the actual calculations: $\newline$ $(x^2 + 3)^2 = x^4 + 2(3x^2) + 9$ $\newline$ This simplifies to: $\newline$ $x^4 + 6x^2 + 9$
Equate to right side of equation: Next, we equate the expanded form to the right side of the original equation: $x^4 + 6x^2 + 9 = 4x^2 + 12$
Move terms to one side: Now, we need to move all terms to one side to set the equation to zero and solve for $x$ : $x^4 + 6x^2 + 9 - 4x^2 - 12 = 0$
Combine like terms: Simplify the equation by combining like terms: $x^4 + 2x^2 - 3 = 0$
Substitute $y = x^2$ : This is a quadratic equation in terms of $x^2$ . Let's substitute $y = x^2$ to make it easier to solve: $\newline$ $y^2 + 2y - 3 = 0$
Factor the quadratic equation: Now, we factor the quadratic equation: $\newline$ $(y + 3)(y - 1) = 0$
Solve for y: Set each factor equal to zero and solve for y: $\newline$ $y + 3 = 0$ or $y - 1 = 0$
Substitute back to find $x$ : Solving for $y$ gives us two solutions: $y = -3$ or $y = 1$
Solve for x: Since $y = x^2$ , we substitute back to find the values of $x$ : $\newline$ $x^2 = -3$ (This is not possible since $x^2$ cannot be negative for real numbers.) $\newline$ $x^2 = 1$
Solve for x: Since $y = x^2$ , we substitute back to find the values of $x$ : $\newline$ $x^2 = -3$ (This is not possible since $x^2$ cannot be negative for real numbers.) $\newline$ $x^2 = 1$ Solve for $x$ when $x^2 = 1$ : $\newline$ $x = \sqrt{1}$ or $x = -\sqrt{1}$ $\newline$ $x = 1$ or $x$ $0$