What are the solutions of the equation x4 – 5×2 – 36 = 0

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

The Correct Answer and Explanation is :

The given equation is:

[
x^4 – 5x^2 – 36 = 0
]

To solve this equation by factoring, follow these steps:

Step 1: Make a substitution

Notice that the equation contains (x^4) and (x^2), which suggests that substituting a new variable might simplify the equation. Let:

[
y = x^2
]

This transforms the equation into:

[
y^2 – 5y – 36 = 0
]

Now, you have a quadratic equation in terms of (y).

Step 2: Factor the quadratic equation

We need to factor the quadratic equation (y^2 – 5y – 36 = 0). To factor, we look for two numbers that multiply to give (-36) (the constant term) and add to give (-5) (the coefficient of (y)).

The two numbers that satisfy this condition are (-9) and (4) because:

[
-9 \times 4 = -36 \quad \text{and} \quad -9 + 4 = -5
]

So, we can factor the quadratic as:

[
(y – 9)(y + 4) = 0
]

Step 3: Solve for (y)

Now that we have the factored form, set each factor equal to zero:

[
y – 9 = 0 \quad \text{or} \quad y + 4 = 0
]

Solving these equations gives:

[
y = 9 \quad \text{or} \quad y = -4
]

Step 4: Substitute back for (x)

Recall that (y = x^2), so we now substitute back:

  1. If (y = 9), then:

[
x^2 = 9
]

Taking the square root of both sides:

[
x = \pm 3
]

  1. If (y = -4), then:

[
x^2 = -4
]

This has no real solutions because the square of a real number cannot be negative. Thus, there are no real solutions from this equation.

Final Answer

The real solutions to the original equation are:

[
x = 3 \quad \text{or} \quad x = -3
]

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