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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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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.

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