Extracting the common factor in parentheses

To solve the equation $x^4 + 2x^2 = 0$, we will use the technique of factoring. Let's proceed step-by-step:

First, notice that both terms $x^4$ and $2x^2$ have a common factor of $x^2$. We can factor $x^2$ out from the equation:

$x^2(x^2 + 2) = 0$

Now, to solve for $x$, we apply the Zero Product Property, which gives us that at least one of the factors must be zero:

• $x^2 = 0$ or
• $x^2 + 2 = 0$

Solving the first case, $x^2 = 0$:

$x = 0$

For the second case, $x^2 + 2 = 0$, we reach:

$x^2 = -2$

Since $x^2 = -2$ has no real solutions (squares of real numbers are non-negative), we can conclude that this equation doesn't provide additional real solutions.

Therefore, the only real solution to the given equation is $x = 0$.

The correct choice from the provided options is:

$x=0$

Shown below is the given equation:

$x^2-x=0$

First note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is $x$and this is due to the fact that the first power is the lowest power in the equation. Therefore it is included both in the term with the second power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest. Hence this is the term with the highest power that can be factored out as a common factor from all terms in the expression. Continue to factor the expression:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0$

Proceed to the left side of the equation that we obtained in the last step. There is a multiplication of algebraic expressions and on the right side the number 0. Therefore given that the only way to obtain 0 from a multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$\boxed{x=0}$

or:

$x-1=0\\ \downarrow\\ \boxed{x=1}$

Let's summarize then the solution to the equation:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0 \\ \downarrow\\ x=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore the correct answer is answer B.

Solve the given equation:

$7x^3-x^2=0$

Note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is $x^2$since the second power is the lowest power in the equation and therefore is included both in the term with the third power and in the term with the second power. Any power higher than this is not included in the term with the second power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression.

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0$

Note that the left side of the equation that we obtained in the last step is a multiplication of algebraic expressions and on the right side the number 0.

Therefore, given that the only way to obtain 0 from a multiplication operation is to multiply by 0. Hence at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^2=0 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(in this case taking the even root of the right side of the equation will indeed yield two possibilities, positive and negative. However since we're dealing with zero, we'll get only one possibility)

or:

$7x-1=0$Let's solve this equation in order to obtain the additional solutions (if they exist) to the given equation:

We obtained a simple first-degree equation which we'll solve by isolating the unknown on one side, we'll do this by moving terms and then dividing both sides of the equation by the coefficient of the unknown:

$7x-1=0 \\ 7x=1\hspace{8pt}\text{/}:7\\ \boxed{x=\frac{1}{7}}$

Let's summarize the solution of the equation:

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0\\ \downarrow\\ x^2=0 \rightarrow\boxed{ x=0}\\ 7x-1=0\rightarrow \boxed{x=\frac{1}{7}}\\ \downarrow\\ \boxed{x=0,\frac{1}{7}}$

Therefore the correct answer is answer C.

Shown below is the given equation:

$7x^{10}-14x^9=0$

First, note that on the left side we are able to factor the expression using a common factor.

The largest common factor for the numbers and variables in this case is $7x^9$given that the ninth power is the lowest power in the equation and therefore is included in both the term with the tenth power and the term with the ninth power. Any power higher than this is not included in the term with the ninth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0$

On the left side of the equation that we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to obtain a result of 0 from a multiplication operation is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^9=0 \hspace{8pt}\text{/}:7\\ x^9=0 \hspace{8pt}\text{/}\sqrt[9]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we proceeded to extract a ninth root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-2=0 \\ \boxed{x=2}$

Let's summarize the solution of the equation:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0\\ \downarrow\\ 7x^9=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore, the correct answer is answer A.

The problem at hand is to solve the equation $x^4 + x^2 = 0$.

Let's begin by factoring the expression:

The given equation is: $x^4 + x^2 = 0$

We can factor out the common factor of $x^2$ from both terms:

$x^2(x^2 + 1) = 0$

To solve for $x$, we set each factor equal to zero:

• $x^2 = 0$

Solving for $x$, we have:

$x = 0$

Next, consider the second factor:

• $x^2 + 1 = 0$

Solving for $x$, we have:

$x^2 = -1$

Since $x^2 = -1$ has no real solutions, we ignore these solutions in the real number system.

Thus, the only real solution to the equation $x^4 + x^2 = 0$ is:

$x = 0$