Question

If $c > 0$ and $b > c$, then ${x^2} + bx - c = 0$ will have:
A) Exactly one root between $0$ and $1$
B) Both root between $0$ and $1$
C) No root between $0$ and $1$
D) None of these

Answer 1

Hint: To solve this problem we have to first use the quadratic formula to find the roots of the given quadratic equation, which is, for a quadratic equation of the form $a{x^2} + bx + c = 0$, the quadratic equation gives,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
On finding the roots of the quadratic equations, we have to use the initial conditions given to us in the question and hence deduce the required answer that how many roots of the quadratic equation are there in between $0$ and $1$, as in the question we can see all three options asks that how many roots are there in between $0$ and $1$.

Complete step by step answer:
The given conditions are,
$c > 0$ and $b > c$.
The quadratic equation is,
${x^2} + bx - c = 0$
Now, using the quadratic formula that is, for a quadratic equation of the form $a{x^2} + bx + c = 0$, the quadratic equation gives,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, from comparing the two quadratic equations, ${x^2} + bx - c = 0$ and $a{x^2} + bx + c = 0$, we can say,
$1 = a,b = b,c = - c$
So, using the quadratic formula on the equation ${x^2} + bx - c = 0$, we get,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4\left( 1 \right)\left( { - c} \right)} }}{{2\left( 1 \right)}}$
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} + 4c} }}{2}$
So, our two roots can be written as,
$x = \dfrac{{ - b + \sqrt {{b^2} + 4c} }}{2}$
And, $x = \dfrac{{ - b - \sqrt {{b^2} + 4c} }}{2}$
Now, taking the first root, $x = \dfrac{{ - b + \sqrt {{b^2} + 4c} }}{2}$, we can say,
Clearly, ${b^2} > b$
$ \Rightarrow {b^2} + 4c > b$
[Since, we are adding a positive number $4c$ to ${b^2}$ and ${b^2} > b$]
$ \Rightarrow \sqrt {{b^2} + 4c} > b$
[Since, ${b^2} + 4c > {b^2} > \sqrt {{b^2}} = b$]
Therefore, we can say, $\sqrt {{b^2} + 4c} - b > 0$.
Therefore, we can say that, $x = \dfrac{{ - b + \sqrt {{b^2} + 4c} }}{2} > 0$.
In the options, we can see that all three options ask for roots in between $0$ and $1$.
So, we can say that, this positive root of the quadratic equation is in between $0$ and $1$.
Now, taking the second root, $x = \dfrac{{ - b - \sqrt {{b^2} + 4c} }}{2}$.
Taking negation common from the numerator, we get,
$ \Rightarrow x = \dfrac{{ - (b + \sqrt {{b^2} + 4c} )}}{2}$
In the numerator, the term within the bracket is positive, as,
$\sqrt {{b^2} + 4c} > 0$
And, $b > c$ and $c > 0$, and hence,
$b > 0$
So, $\sqrt {{b^2} + 4c} + b > 0$
Now, multiplying both sides of the inequality with $ - 1$, we get,
$ - \left( {\sqrt {{b^2} + 4c} + b} \right) < 0$
[Since, the direction of an inequality changes on taking negation across the inequality]
So, we can say, that,
$x = \dfrac{{ - (b + \sqrt {{b^2} + 4c} )}}{2} < 0$
Therefore, the second root of the quadratic equation is less than $0$.
So, we can conclude that, only one root of the quadratic equation is between $0$ and $1$.
Therefore, the correct option is option (A).

Note:
We can observe a lot about the answer from analysing the options of an objective question. We can also conclude about an answer including the quadratic equation by finding its discriminant $D$, which is $D = {b^2} - 4ac$. If $D > 0$, the roots are real and distinct, if $D = 0$, the roots are real and equal and, if $D < 0$, then the roots are imaginary and non-real.