Roots Lie In An Interval

ROOTS LIE IN AN INTERVAL

Here basically we will discuss different necessary and sufficient conditions we should impose on a quadratic equation ax² + bx + c = 0 such that roots of the given equation lies in a particular interval. Since a $\ne$ 0 , we can take f(x) = x² + $\frac{b}{a}x+\frac{c}{a}$

Case I: Both the roots are positive i.e. they lie in (0, $\infty$), then the sum of the roots as well as the product of the roots must be positive.

$\Rightarrow$ $\alpha$ + $\beta$ = -$\frac{b}{a}>0$ and $\alpha$$\beta$ = $\frac{c}{a}>0$ with b² – 4ac $\ge$ 0.

Case II: Both the roots are negative i.e. they lie in (- $\infty$, 0), then the sum of the roots must be negative and the product of the roots must be positive. i.e. $\alpha$ + $\beta$ = -$\frac{b}{a}$ < 0 and $\alpha $$\beta $ = $\dfrac{c}{a}>0$ with b² – 4ac $\ge$ 0.

Case III: One root is positive and other is negative i.e. origin is lying between the roots. Clearly f(0)<0 is the necessary and sufficient condition.

Case IV: Both the roots are greater then a real number k.

D $\ge$ 0 … (1)

f(k) > 0 …(2)

$-\frac{b}{2a}$ > k …(3)

These are the necessary & sufficient conditions.

Case V: If both the roots are less than a real number k.

D $\ge$ 0 … (1)

f(k) > 0 …(2)

$-\frac{b}{2a}$< k …(3)

These are the necessary & sufficient conditions.

Case VI: A real number k is lying between the roots i.e. one root is less then k and other is greater then k.

D > 0 … (1)

f(k) < 0 … (2)

These are the necessary & sufficient conditions.

Case VII: exactly are root is lying between k₁ and k₂

f(k₁) < 0 and f(k₂) > 0 f(k₁) > 0 and f(k₂) < 0

Hence the required condition is f(k₁). f(k₂) < 0

Illustration 1 : For the quadratic equation x² – (m – 3)x + m = 0, find the value of m for which

(i) one root is smaller than 2 and the other is greater than 2

(ii) both roots are grater than 2

(iii) both roots lie in (1, 2)

(iv) exactly one root lie in (1, 2)

Solution: Let f(x) = x² – (m – 3)x + m and D is = (m – 1)(m – 9)

(i) (a) D> 0 and (b) f (2) < 0

i.e., m < 1 or m > 9 and, m > 10

$\therefore$ m $\in$ (10, $\infty$)

(ii) The required necessary and sufficient conditions are

D $\ge$ 0$\Rightarrow$ m $\le$ 1 or m $\ge$ 9………(1)

f(2) > 0 $\Rightarrow$ m < 10………..(2)

$-\frac{b}{2a}$ > 2$\Rightarrow$ m > 7……………(3)

From (1),(2) and (3) m $\in$ [9, 10)

(iii) D $\ge$ 0$\Rightarrow$ m$\le$1 or m$\ge$9 ……(1)

af(1) > 0$\Rightarrow$ 4>0 $\forall$m$\in$R ……(2)

af(2) > 0$\Rightarrow$ m < 10 ……(3)

1 < $\frac{\alpha +\beta }{2}$ < 2$\Rightarrow$ m>5 and m<7 ……(4)

Taking intersection of these four conditions , we get m $\in$ $\varphi$.

(iv) D > 0 $\Rightarrow$ m < 1 or m > 9 ……(1)

f(1) . f(2) < 0$\Rightarrow$ m > 10 ……(2)

Taking intersection of these two conditions, we get m $\in$ (10, $\infty$)