Question

If the roots of \[a{x^2} - bx - c = 0\] changes by the same quantity, then the expression in a, b, c that does not change is
A) \[\dfrac{{{b^2} - 4ac}}{{{a^2}}}\]
B) \[\dfrac{{{b^2} - 4c}}{a}\]
C) \[\dfrac{{{b^2} + 4ac}}{{{a^2}}}\]
D) None of these

Answer 1

Hint: At first, we will find the roots of the given equation. Next, we apply a change on the roots and we will get a new quadratic equation. There, we will get the new roots.
The, we can compare the discriminant of both the equation and we can find if there is any change.

Complete step-by-step answer:
It is given that; the roots of \[a{x^2} - bx - c = 0\]…………... (1) changes by the same quantity.
We have to find out the given problem.
Let us consider a quadratic equation \[a{x^2} + bx + c = 0,a \ne 0\]. The roots of this equation are \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] . And, the discriminant of this equation is, \[{b^2} - 4ac\].
The roots of the equation (1) are \[\dfrac{{b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\].
Let us take the roots of \[a{x^2} - bx - c = 0\] are \[\alpha ,\beta \].
Let us increase the roots by some constant k.
Therefore, the roots become, \[\alpha + k,\beta + k\].
Now, \[x = \alpha \] satisfies the above equation.
Let, \[\alpha + k = x \Rightarrow \alpha = x - k\]
Substitute the value of \[\alpha \]in the above equation we get,
\[a{(x - k)^2} - b(x - k) - c = 0\]
Simplifying we get,
\[{x^2}(a) - x(2ak + b) + a{k^2} + bk - c = 0\]..............… (2)
Now, we are going to check the discriminant of the equation (2).
So, we have,
\[D = \dfrac{{(2ak + b) \pm \sqrt {4{a^2}{k^2} + {b^2} + 4abk - 4{a^2}{k^2} - 4abk + 4ac} }}{{2a}}\]
Simplifying we get,
\[D = \dfrac{{(2ak + b) \pm \sqrt {{b^2} + 4ac} }}{{2a}}\]
We can see the discriminant quantity is the same in both the equation (1) and (2).

$\therefore $The correct option is D) none of these.

Note: In this question, we found the expression that in a, b, c that does not change. Students must focus on the steps that check the discriminant for the roots. Because on that step we substituted the roots values into the relation.
If Discriminant is less than 0 then the roots are imaginary.
If the discriminant is equal to 0 then the roots are equal and rational.
If the discriminant is greater than 0 then the unequal.