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Does there exist a quadratic equation whose coefficients are all distinct irrationals but both the roots are rational? Why?

Answers (1)

True

Let us consider an equation with distinct irrational numbers

 \sqrt{2}x^{2}-5\sqrt{2}x+4\sqrt{2}=0

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

a=\sqrt{2},b=-5\sqrt{2},c=4\sqrt{2}

let us find the roots of the equation

 \\x=\frac{-b+\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{5\sqrt{2 }\pm \sqrt{50-4(\sqrt{2})(4\sqrt{2})}}{2 \sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{50-32}}{2\sqrt{2}}\\ x=\frac{5 \sqrt{2} \pm \sqrt{18}}{2\sqrt{2}}\\ x=\frac{\sqrt{2}(5\pm 3)}{2\sqrt{2}}=\frac{5 \pm 3}{2}\\ x=\frac{5+3}{2}=4 \; \; \; \; \; \; \; \; x=\frac{5-3}{2}=1

The roots are rational. Hence the given statement is true

           

Posted by

infoexpert24

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