KCET · Maths · Determinants
If \( \alpha \) and \( \beta \) are the roots of \( x^{2}-a x+b^{2}=0 \), then \( \alpha^{2}+\beta^{2} \) is equal to
- A \( a^{2}-2 b^{2} \)
- B \( 2 a^{2}-b^{2} \)
- C \( a^{2}-b^{2} \)
- D \( a^{2}+b^{2} \)
Answer & Solution
Correct Answer
(A) \( a^{2}-2 b^{2} \)
Step-by-step Solution
Detailed explanation
Given equation,
\(x^{2}-a x+b^{2}=0 \rightarrow(1)\)
We know that, sum of the roots is given by
\(\alpha+\beta=a\)
and the product of the roots is given by
\(\alpha \beta=b^{2}\)
Now, \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta\)
\(=a^{2}-2 b^{2}\)
\(x^{2}-a x+b^{2}=0 \rightarrow(1)\)
We know that, sum of the roots is given by
\(\alpha+\beta=a\)
and the product of the roots is given by
\(\alpha \beta=b^{2}\)
Now, \(\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta\)
\(=a^{2}-2 b^{2}\)
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