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(C) Given that the quadratic equation has rational coefficients and one root is $\sqrt{5}$.Since irrational roots of a quadratic equation with rationa

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$x^{2}-5=0$

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(C) Given that the quadratic equation has rational coefficients and one root is $\sqrt{5}$.Since irrational roots of a quadratic equation with rational coefficients always occur in conjugate pairs,the other root must be $-\sqrt{5}$.Sum of the roots $= \sqrt{5} + (-\sqrt{5}) = 0$.Product of the roots $= (\sqrt{5}) 	imes (-\sqrt{5}) = -5$.The standard form of a quadratic equation is $x^{2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^{2} - (0)x + (-5) = 0$.Therefore,the required equation is $x^{2} - 5 = 0$.

The quadratic equation with rational coefficients,whose one root is $\sqrt{5}$,is:

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