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We know that if the discriminant of a quadratic equation is greater than zero & it is a perfect square then roots are real, distinct & rational. I was solving a question in which the quadratic equation had rational coefficients & when I found its discriminant, it was a perfect square but still I found that its roots are irrational...is it possible as I know that quadratic equations irrational roots & rational coefficients...always occur in conjugate pairs....BUT the problem was that my D was prect square & still roots were irrational. This is the question 2x 2 - 2 root 3x + 1 We know that if the discriminant of a quadratic equation is greater than zero & it is a perfect square then roots are real, distinct & rational. I was solving a question in which the quadratic equation had rational coefficients & when I found its discriminant, it was a perfect square but still I found that its roots are irrational...is it possible as I know that quadratic equations irrational roots & rational coefficients...always occur in conjugate pairs....BUT the problem was that my D was prect square & still roots were irrational. This is the question 2x2- 2 root 3x + 1
We know that if the discriminant of a quadratic equation is greater than zero & it is a perfect square then roots are real, distinct & rational. I was solving a question in which the quadratic equation had rational coefficients & when I found its discriminant, it was a perfect square but still I found that its roots are irrational...is it possible as I know that quadratic equations irrational roots & rational coefficients...always occur in conjugate pairs....BUT the problem was that my D was prect square & still roots were irrational. This is the question 2x2- 2 root 3x + 1
roots of 2x^2 +2 are 1 and -1 not -1/3 and when D is perfect square and rational cofficients are there than roots are always real and rational
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