Question

1. state the factors of -36 that add up to -5 in the problem below.

$x^4 - 5x^2 - 36 = 0$
$-9, 4$
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1. what is the new equation after step 1?

$x^4 - 5x^2 - 36 = 0$

Explanation:

Step1: Let \( y = x^2 \)

Substitute \( y = x^2 \) into the equation \( x^4 - 5x^2 - 36 = 0 \). Since \( x^4=(x^2)^2 = y^2 \), the equation becomes \( y^2 - 5y - 36 = 0 \).

Step2: Factor the quadratic (using the factors from problem 3)

From problem 3, we know the factors of - 36 that add up to - 5 are - 9 and 4. So we can factor \( y^2 - 5y - 36 \) as \( (y - 9)(y+4)=0 \) (because \( y^2-5y - 36=y^2+4y-9y - 36=y(y + 4)-9(y + 4)=(y - 9)(y + 4) \)). Then substitute back \( y=x^2 \), we get \( (x^2 - 9)(x^2+4)=0 \). But if we just consider the first step of substitution (step 1's result), the new equation after substituting \( y = x^2 \) is \( y^2-5y - 36 = 0 \) or in terms of factoring with the known factors, \( (y - 9)(y + 4)=0 \) (substituting back \( y=x^2 \) gives \( (x^2 - 9)(x^2 + 4)=0 \)). However, if step 1 is substitution, the new equation is \( y^2-5y - 36 = 0 \) where \( y=x^2 \), or directly factored as \( (x^2 - 9)(x^2 + 4)=0 \). But the most common first step for solving a bi - quadratic equation like \( x^4-5x^2 - 36 = 0 \) is substitution \( y = x^2 \), so the new equation is \( y^2-5y - 36 = 0 \) or after factoring (using the factors from part 3) \( (x^2 - 9)(x^2 + 4)=0 \). If we follow the factoring from part 3, the factored form is \( (x^2 - 9)(x^2 + 4)=0 \) (since we use the factors - 9 and 4 to factor the quadratic in \( y \) and then substitute back).

Answer:

If step 1 is substitution (\( y=x^2 \)): \( y^2 - 5y - 36=0 \) (or \( (x^2 - 9)(x^2 + 4)=0 \) after factoring)