Does there exist a quadratic equation whose coefficients are rationals but both of its roots are irrational? why ?
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yes i guess.
well,the irrational roots depends on the underroot part of the fomula and the underroot wont remove unless its a perfect square.
To find the root of the quadratic eqn,we use the formula:
x=[a+-sqrt(b^2-4ac)]/2*a
In this eqn:b^2-4ac is the term which decides whether the root will be irrational or not.
If,for example,the value of this term comes out to be 4 or 16 or any perfect square then the root part will be removed and the roots will become rational.
BUT if this term comes out to be 2 or 5 or some number that isnt a perfect square then the roots will be irrational.
So answer to ur qn is yes,its possible to get irrational roots with rational coefficients.
well,the irrational roots depends on the underroot part of the fomula and the underroot wont remove unless its a perfect square.
To find the root of the quadratic eqn,we use the formula:
x=[a+-sqrt(b^2-4ac)]/2*a
In this eqn:b^2-4ac is the term which decides whether the root will be irrational or not.
If,for example,the value of this term comes out to be 4 or 16 or any perfect square then the root part will be removed and the roots will become rational.
BUT if this term comes out to be 2 or 5 or some number that isnt a perfect square then the roots will be irrational.
So answer to ur qn is yes,its possible to get irrational roots with rational coefficients.
garimavirodhiya:
thanks but not satisfy
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