Question
\( \left(b^{2}-3\right)^{2}+\left(b^{2}-3\right)-42=0 \) The solution set is \( \{\square\} \).
Ask by Davies Cervantes. in the United States
Feb 25,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The solutions are \( b = -3 \) and \( b = 3 \).
Solution
Solve the equation \( (b^{2}-3)^{2}+(b^{2}-3)-42=0 \).
Solve the equation by following steps:
- step0: Solve for \(b\):
\(\left(b^{2}-3\right)^{2}+\left(b^{2}-3\right)-42=0\)
- step1: Simplify:
\(\left(b^{2}-3\right)^{2}+b^{2}-45=0\)
- step2: Calculate:
\(b^{4}-5b^{2}-36=0\)
- step3: Factor the expression:
\(\left(b-3\right)\left(b+3\right)\left(b^{2}+4\right)=0\)
- step4: Separate into possible cases:
\(\begin{align}&b-3=0\\&b+3=0\\&b^{2}+4=0\end{align}\)
- step5: Solve the equation:
\(\begin{align}&b=3\\&b=-3\\&b \notin \mathbb{R}\end{align}\)
- step6: Find the union:
\(\begin{align}&b=3\\&b=-3\end{align}\)
- step7: Rewrite:
\(b_{1}=-3,b_{2}=3\)
The solution set for the equation \( (b^{2}-3)^{2}+(b^{2}-3)-42=0 \) is \( \{b=-3, b=3\} \).
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Simplify this solution Mind Expander
To solve the equation \( \left(b^{2}-3\right)^{2}+\left(b^{2}-3\right)-42=0 \), we can start by letting \( x = b^{2} - 3 \). Substituting this into the equation simplifies it to \( x^{2} + x - 42 = 0 \). Next, we can factor the quadratic as \( (x - 6)(x + 7) = 0 \). This gives us two solutions for \( x \): \( x = 6 \) and \( x = -7 \). Substituting back for \( x \), we get \( b^{2} - 3 = 6 \) leading to \( b^2 = 9 \) which gives \( b = 3 \) or \( b = -3 \). For \( x = -7 \), we have \( b^{2} - 3 = -7 \), resulting in \( b^{2} = -4 \) which has no real solutions. Thus, the solution set is \( \{3, -3\} \).
