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Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational? Justify your answer.

Answers (2)

True

Let us suppose a quadratic equation with rational coefficients.

x^{2}-2x+4=0

compare with ax^{2} + bx + c = 0 where a \neq 0

a=-1,b=-2,c=4

let us find the roots of the equation

 \\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(-1)(4)}}{2 (-1)}\\ x=\frac{2 \pm \sqrt{20}}{-2}\\ x=\frac{2\pm 2\sqrt{5}}{-2}\\ x=\frac{2+2\sqrt{5}}{-2} \; \; \; \; \; \; \; \; x=\frac{2-2\sqrt{5}}{-2}\\

Both of its roots are irrational

Hence the given statement is true

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infoexpert24

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True

Solution

            Rational number – A number that can be expressed in the form of \frac{p}{q}  where q\neq 0

            (p, q are integers)

            Irrational number – A number that can not be expressed in the form of ratio of two integers.

            Let us suppose a quadratic equation with rational coefficient.

                     -x^{2}-2x+4=0

               compare with ax^{2} + bx + c = 0 where a \neq 0

              Let us find the roots of the equation

                       x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(-1)(4)}}{2 (-1)}\\ x=\frac{2 \pm \sqrt{20}}{-2}\\ x=\frac{2\pm2\sqrt{5}}{-2}\\ x=\frac{2+2\sqrt{5}}{-2} \; \; \; \; \; \; \; \; x=\frac{2-2\sqrt{5}}{-2}\\

            Both of its roots are irrational.

            Hence the given statement is true.

Posted by

infoexpert24

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