Quadratic Equation

Q: For what values of the parameter a does the equation x⁴ + 2ax³ + x² + 2ax + 1 = 0, have atleast two distinct negative roots .

Sol. Given equation

x⁴ + 2ax³ + x² + 2ax + 1 = 0

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

solution of equation (1) can be find out by finding the intersection of the curve y = a

and $\displaystyle y = -\frac{x^4 + x^2 + 1}{2(x^3 + 2x)} $

for x < 0, Line y = a will cut the curve in two distinct point, if a > 3/4 .

Q: Show that the equation $\displaystyle \frac{A^2}{(x-a)} + \frac{B^2}{(x-b)} + \frac{C^2}{(x-c)} + …..+ \frac{H^2}{(x-h)} = k $ has no imaginary root, where A, B, C …. ,H and a, b, c ….., h and k ∈ R.

Sol: Suppose one root of the equation is (u +iv) then other root would be u – iv

$\displaystyle \frac{A^2}{(u-a)+ i v} + \frac{B^2}{(u-b)+ i v} + \frac{C^2}{(u-c)+ i v} + …..+ \frac{H^2}{(u-h)+ i v} = k $ …(i)

$\displaystyle \frac{A^2}{(u-a)- i v} + \frac{B^2}{(u-b)- i v} + \frac{C^2}{(u-c)- i v} + …..+ \frac{H^2}{(u-h)- i v} = k $ …(ii)

(i)-(ii) we get ,

$\displaystyle iv[\frac{A^2}{(u-a)^2 + v^2} + \frac{B^2}{(u-b)^2 + v^2} + \frac{C^2}{(u-c)^2 + v^2} + …..+ \frac{H^2}{(u-h)^2 +v^2}] = 0 $

This is possible only when v = 0 and for this case there is no imaginary root.

Q: Find the values of ‘a ‘ for which the equation
(x² + x + 2)² – (a – 3) (x² + x + 2) (x² + x + 1) +( a – 4) (x² + x +1)² = 0 has at least one real root.

Sol: Putting x² + x +1 = α ,

we get (α +1)² – ( a – 3)α( α +1) + ( a – 4) α² = 0

⇒ α (5 – a ) + 1 = 0

⇒ α =-1/(5-a)

⇒ x² + x + 1 =1/(a-5)

$\displaystyle \frac{1}{a-5} \ge \frac{3}{4}$

⇒ a – 5 > 0 and a – 5 ≤ 4/3

⇒ 5 < a  ≤ 19/3.

Q: Let a , b , c be real. If ax^2 + bx + c = 0 has two real roots α and β where α < -1 and β > 1, then show that $\displaystyle 1 + \frac{c}{a} + |\frac{b}{a}| < 0 $

Sol: Let $\displaystyle f(x) = x^2 + \frac{b}{a} x + \frac{c}{a} $

On drawing graph f(-1) < 0 and f(1) < 0

$\displaystyle 1 + \frac{c}{a} – \frac{b}{a} < 0 $

and , $\displaystyle 1 + \frac{c}{a} + \frac{b}{a} < 0 $

⇒ $\displaystyle 1 + \frac{c}{a} + |\frac{b}{a}| < 0 $