If (x^2-2(a+b)x+2ab=0) has real roots, which statement is always true for (a) and (b)?
Answer and explanation
Correct answer: Roots are always real because (a^2+b^2\geq0)
For the equation \\(x^2-2(a+b)x+2ab=0\\), the coefficients are 1, \\(-2(a+b)\\) and 2ab. Its discriminant is \\(D=[-2(a+b)]^2-8ab=4(a+b)^2-8ab=4(a^2+b^2)\\). Since squares are non-negative, \\(a^2+b^2\geq0\\), so D is always non-negative.
Therefore the equation always has real roots, including the possibility of equal roots when both a and b are zero. Option A expresses this correctly. The condition ab>0 is not necessary; a and b may have opposite signs or one may be zero. Also, equal roots require D=0, namely a=b=0, not merely a+b=0.
Frequently asked questions
What is the correct answer to this question?
Roots are always real because (a^2+b^2\geq0)
Why is this the correct answer?
For the equation \\(x^2-2(a+b)x+2ab=0\\), the coefficients are 1, \\(-2(a+b)\\) and 2ab. Its discriminant is \\(D=[-2(a+b)]^2-8ab=4(a+b)^2-8ab=4(a^2+b^2)\\). Since squares are non-negative, \\(a^2+b^2\geq0\\), so D is always non-negative.
Therefore the equation always has real roots, including the possibility of equal roots when both a and b are zero. Option A expresses this correctly. The condition ab>0 is not necessary; a and b may have opposite signs or one may be zero. Also, equal roots require D=0, namely a=b=0, not merely a+b=0.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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