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Roots of the equation `(x^2-4x+3)+lamda(x^2-6x+8)=0` , `lamda in R` will beA. always realB. real only when `lambda` is positiveC. real only when `lambda` is negativeD. always is magnary |
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Answer» Correct Answer - A Roots of the quadratic ………….. `(x^(2)-4x+3)+lambda(x^(2)-6x+8)=0` `x^(2)(1+lambda)-2x(2+3lambda)+(3+8lambda) = 0` Discriminant. `D = 4(2+3lambda)^(2)-4(1+lambda)(3+8lambda)` `D = 4(lambda^(2)+lambda+1)` If `lambda epsilon R` then `D gt 0` so root of given quadratic always real |
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