Let ax^{2} + bx + c = 0 and dx^{2} + ex + f = 0 have a common root α(say).

Then aα^{2} + bα + c = 0

and dα^{2}+ eα + f = 0.

Solving for α^{2} 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 x^{2} + cx + ab = 0 and x^{2} + bx + ca = 0 have a common root, then show that their other roots satisfy the equation x^{2} + ax + bc = 0.

Solution:

Let the equation x^{2} + cx + ab = 0 ….(1)

and x^{2} + bx + ca = 0 ….(2)

have the common root α

Therefore, α^{2} + cα + ab = 0 and α^{2} + 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 β_{1} respectively then α.β = ab and αβ_{1} = ca

But α = a.

Therefore β = b , β_{1} = c

Now the equation whose roots are β and β_{1} is

x^{2} − (β + β_{1})x + β.β_{1} = 0

=> x^{2} − (b + c)x + bc = 0

=> x^{2} + ax + bc = 0 (Since b + c = -a)

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

x^{2} + 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 x_{1} and x_{2} (x_{1} < x_{2}) 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 α_{1} , α_{2} , ….., α_{n} , β_{1}, β_{2} , …., β_{m} are distinct real numbers.

Then f(x) = 0 for x = α_{1} , α_{2} , …., α_{n},

and f(x) is not defined for x = β_{1} ,β_{2} , ……, β_{m}

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

Now arrange α_{1} , α_{2} , …., α_{n} , β_{1} , β_{2} , ……, β_{m} in an increasing order say c_{1} , c_{2} , ……, c_{m+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 ax^{2} + bx + c = 0 and x^{3} + 3x^{2} + 3x + 2 = 0 have two common roots, then show that a = b = c