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Does there exist a quadratic equation whose co-efficient are all distinct irrationals . . .

Question :
a) Does there exist a quadratic equation whose co-efficient are all distinct irrationals but both the roots are rational? Why?
b) Does there exist a quadratic equation whose co-efficient are rational but both of its roots are irrational? Why?

Doubt by Ananya

Solution :

a) Let us consider a quadratic equation
ax2+bx+c=0
where a, b and c are coefficients of x2, x and constant term respectively.

Let α and β are the roots of the quadratic equation.

α+β = -b/a &
αβ = c/a

where a, b and c are irrational numbers.

We know, when an irrational number is divided by another irrational number then the answer may be rational or irrational.
[If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational.]

So, we can say that there could be a quadratic equation whose co-efficient are all distinct irrationals but both the roots are rational.

Some examples : √3x2–7√3x+12√3 = 0 
Here all the coefficients are irrational but the roots are 3 and 4 which are rational.

b) Let us consider a quadratic equation
ax2+bx+c=0
where a, b and c are coefficients of x2, x and constant term respectively.

Let α and β are the roots of the quadratic equation.

α+β = -b/a &
αβ = c/a

where a, b and c are rational numbers.


We know, sum or product of two irrational numbers (α & β) could give both rational and irrational.
[If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational.] 

So, we can say that there could be a quadratic equation whose co-efficient are all rational but both the roots are irrational.

Some examples : x2-3x+1=0 
Here all the coefficients are rational but the roots are (3+√5)/2 and (3-√5)/2  which are irrational.