**What are the solutions of the equation x4 – 5×2 – 36 = 0? use factoring to solve.**

- $x=−3, 3$
- $x=−4,9$
- $x=4,−9$
- $x=0$

### The Correct Answer is

$a. x=−3, 3$

**Correct Answer Explanation: $a. x=−3, 3$**

Let’s solve the given quadratic equation $x_{4}−5x_{2}−36=0$ using factoring.

**Step 1: Recognizing the Quadratic Form**

Notice that this equation is in quadratic form, meaning we can represent it as $(x_{2}_{2}−5(x_{2})−36=0$.

**Step 2: Factoring**

Now, let’s introduce a substitution to make this equation easier to factor. Let $y=x_{2}$, so our equation becomes $y_{2}−5y−36=0$.

Now, we need to factor the quadratic expression $y_{2}−5y−36$. We are looking for two numbers whose product is $−36×1$ (coefficient of $y_{2}×constant term$) and whose sum is $−5$ (coefficient of $y$).

The numbers that satisfy this condition are $−9$ and $4$ because $−9×4=−36$ and $−9+4=−5$.

So, we can express $y_{2}−5y−36$ as $(y−9)(y+4)=0$.

**Step 3: Substitute Back**

Now, substitute back $x_{2}$ for $y$: $(x_{2}−9)(x_{2}+4)=0$.

Now, we have a product of two factors equal to zero. According to the zero-product property, this implies that either $x_{2}−9=0$ or $x_{2}+4=0$.

**Solving for $x_{2}−9=0$:**

$x_{2}−9=0$

Add 9 to both sides:

$x_{2}=9$

Take the square root of both sides:

$x=±3$

**Solving for $x_{2}+4=0$:**

$x_{2}+4=0$

Subtract 4 from both sides:

$x_{2}=−4$

This has no real solutions because the square of any real number is non-negative, and here we’re looking for a square to be $−4$.

The solutions to the original equation $x_{4}−5x_{2}−36=0$ are $x=−3,3$ based on the solutions obtained from $x_{2}−9=0$.

**Why the other solutions are Incorrect:**

**b. Incorrect Solution: $x=−4,9$**

This solution is incorrect because it does not satisfy the original equation $x_{4}−5x_{2}−36=0$. If we substitute $x=−4$ or $x=9$ into the original equation, we get $256−5(16)−36$ or $81−5(81)−36$, neither of which equals zero.

**c. Incorrect Solution: $x=4,−9$**

Similarly, this solution is incorrect. Substituting $x=4$ or $x=−9$ into the original equation results in $256−5(16)−36$ or $81−5(81)−36$, which are not equal to zero.

**d. Incorrect Solution: $x=0$**

This solution is also incorrect. When substituting $x=0$ into the original equation, we get $(−36)$, which is not equal to zero.

All the incorrect solutions were evaluated by substituting them into the original equation $x_{4}−5x_{2}−36=0$, and in each case, the equation did not hold true. Therefore, the correct solutions are $x=−3,3$, and the incorrect solutions are $x=−4,9$ and $x=4,−9$, and $x=0$.

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