When faced with the factorization \(\displaystyle{\left(\alpha-{1}-\lambda\right)}{\left(\mu-{1}-\lambda\right)}={0}\) the roots of

2akondupa

2akondupa

Answered question

2022-03-12

When faced with the factorization
(α1λ)(μ1λ)=0
the roots of λ can simply be said to be α1 and μ1.
Why is that answer different to the following working?

Answer & Explanation

Heidy David

Heidy David

Beginner2022-03-13Added 5 answers

Step 1
λ1,2=α+μ2±(αμ)22
=α+μ2±(αμ)2
λ1=2α22=α1
λ2=2μ22=μ1
What seems to be the problem. I cannot be sure how you ended up with the solutions that you did.

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