Question

Prove that the roots of the quadratic equation

( a – b + c ) x^2 + 2 ( a – b ) x + ( a – b – c ) = 0 are rational numbers

for all real numbers a and b and for all rational c.​

Answer 1

the question is from which chapter tell me

Answer 2

Answer:

Let us recall the general solution, α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a

Case I: D^2 = b2 – 4ac > 0  

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive, then the roots α and β of the quadratic equation ax2 +bx+ c = 0 are real and unequal.  

Case II: b2– 4ac = 0  

When a, b, and c are real numbers, a ≠ 0 and discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.  

Case III: b2– 4ac < 0

When a, b, and c are real numbers, a ≠ 0 and discriminant is negative, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are unequal and not real. In this case, we say that the roots are imaginary.  

Case IV: b2 – 4ac > 0 and perfect square

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real, rational and unequal.

Case V: b2– 4ac > 0 and not perfect square  

When a, b, and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax2 + bx + c = 0 are real, irrational and unequal.  

Here the roots α and β form a pair of irrational conjugates.  

Case VI: b2– 4ac >0 is perfect square and a or b is irrational

When a, b, and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax2 + bx + c = 0 are irrational.

Given question is case no (iv) so option B is tru

Step-by-step explanation:

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