# Condition for common root(s) & Method of Intervals

Let ax2 + bx + c = 0 and dx2 + ex + f = 0 have a common root α(say).

Then aα2 + bα + c = 0

and dα2+ eα + f = 0.

Solving for α2 and α , we get

$\displaystyle \frac{\alpha ^2}{bf-ce} = \frac{\alpha}{dc-af} = \frac{1}{ae-bd}$

Hence , $\displaystyle \alpha^2 = \frac{bf-ce}{ae-bd} \, and \, \alpha = \frac{dc-af}{ae-bd}$

⇒ (dc – af)2 = (bf – ce) (ae – bd) ,

which is the required condition for the two equations to have a common root.

Note: Condition for both the roots to be common is

$\displaystyle \frac{a}{d} = \frac{b}{e} = \frac{c}{f}$

Illustration . If x2 + cx + ab = 0 and x2 + bx + ca = 0 have a common root, then show that their other roots satisfy the equation x2 + ax + bc = 0.

Solution:

Let the equation x2 + cx + ab = 0 ….(1)

and x2 + bx + ca = 0    ….(2)

have the common root α

Therefore, α2 + cα + ab = 0 and α2 + bα + ca = 0

$\displaystyle \frac{\alpha^2}{ac^2 -ab^2} = \frac{\alpha}{ab-ac} = \frac{1}{b-c}$

⇒ α = − (b + c) and also α = a

⇒ a + b + c = 0

Let the other root of equations (1) and (2) be β and β1 respectively then α.β = ab and αβ1 = ca

But α = a.

Therefore β = b , β1 = c

Now the equation whose roots are β and β1 is

x2 − (β + β1)x + β.β1 = 0

=> x2 − (b + c)x + bc = 0

=> x2 + ax + bc = 0 (Since b + c = -a)

Hence other roots of equation (1) and (2) are given by the equation

x2 + ax + bc = 0

### Method of Intervals (Wavy Curve Method)

In order to solve inequalities of the form

$\displaystyle \frac{P(x)}{Q(x)}\ge 0 \, , \frac{P(x)}{Q(x)}\le 0$

where P(x) and Q(x) are polynomials, we use the following result:

If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation , then within this interval the polynomial itself takes on values having the same sign.

Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write

$\displaystyle \frac{P(x)}{Q(x)} = f(x) = \frac{(x-\alpha_1)(x-\alpha_2)—(x-\alpha_n)}{(x-\beta_1) (x-\beta_1)—(x-\beta_m)}$

where α1 , α2 , ….., αn , β1, β2 , …., βm are distinct real numbers.

Then f(x) = 0 for x = α1 , α2 , …., αn,

and f(x) is not defined for x = β12 , ……, βm

Apart from these (m + n) real numbers f(x) is either positive or negative.

Now arrange α1 , α2 , …., αn , β1 , β2 , ……, βm in an increasing order say c1 , c2 , ……, cm+n .

Plot them on the real line.

Draw a curve starting from right along the real line which alternately changes its position at these points. This curve is known as the wavy curve. The intervals in which the curve is above the real line will be the intervals for which f(x) is positive and intervals in which the curve is below the real line will be the intervals in which f(x) is negative.

Illustration : Find the set of all x for which

$\displaystyle \frac{2 x}{2 x^2 + 5x +2} > \frac{1}{x + 1}$

Solution: We have

$\displaystyle \frac{2 x}{2 x^2 + 5x +2} – \frac{1}{x + 1} > 0$

$\displaystyle \frac{3x+2}{(x+1)(2x+1)(x+2)} < 0$

$\displaystyle \frac{(3x+2)^2}{(x+1)(2x+1)(x+2)(3x+2)} < 0$

There are five intervals

x < – 2 , – 2 < x < – 1 , – 1< x < – 2/3 , – 2/3 < x < – 1/2 , x > -1/2

The inequality (i) will hold for -2< x < -1 and for -2/3 < x <-1/2

Hence – 2 < x < – 1 and – 2/3 < x < – 1/2

Exercise : Solve the following inequalities:

(i) $\displaystyle \frac{x-1}{x^2 -4x +3 } < 1$

(ii) $\displaystyle \frac{x^2 – 2x-1}{x + 1 } < x$

(iii) $\displaystyle \frac{x-1}{x } – \frac{x+1}{x-1} < 2$

(iv) If the equation ax2 + bx + c = 0 and x3 + 3x2 + 3x + 2 = 0 have two common roots, then show that a = b = c