Quadratic Equation

Q: If roots of the equation x² – 10 c x – 11 d = 0 are a, b and those of x² – 10 a x – 11 b = 0 are c , d , then find the value of a + b + c + d. (a, b, c and d are distinct numbers)

Solution : As a + b = 10 c and c + d = 10 a

ab = -11 d , cd = -11 b

⇒ ac = 121 and (b + d) = 9(a + c)

a² – 10 ac – 11d = 0

c² – 10 ac – 11b = 0

⇒ a² + c² – 20ac – 11(b + d) = 0

⇒ (a + c)² – 22(121) – 11 × 9(a + c) = 0

⇒ (a + c) = 121 or -22 (rejected)

Hence , a + b + c + d = 1210.

Q: If α be a root of the equation 4 x² + 2 x – 1 = 0 , then prove that 4 α ³ – 3α is the other root.

Solution : Let f (x) = 4 x² + 2x – 1

then f (α) = 4α² + 2α – 1 = 0 …(i)

Now 4 α³ – 3α

= 4 α³ .α – 3α

= (1 – 2α)α – 3α from (i)

= α – 2α² – 3α

= –2(α² + α)

$\displaystyle = -2 (\frac{1-2\alpha}{4} + \alpha)$

$\displaystyle = -2 (\frac{1-2\alpha + 4 \alpha}{4} )$

$\displaystyle = – \frac{2 \alpha + 1}{2} = – \alpha – \frac{1}{2}$

Sum of roots = (4 α³ – 3α ) + α = – 1/2

Hence other root is (4 α³ – 3α )

Q: If α is a root of equation 3ax² + 2bx + 6c = 0 and β is a root of equation –3ax² + 2bx + 6c = 0 then prove that the equation 2ax² + 2bx + 6c = 0 has a root γ which lies between a and b .

Solution : α is root of 3ax² + 2bx + 6c = 0

⇒ 3a α² + 2b α + 6c = 0 … (i)

Also β is a root of –3ax² + 2bx + 6c = 0

⇒ -3aβ² + 2b β + 6c = 0 …(ii)

Let f(x) = 2ax² + 2bx + 6c

f(α) = 2a α² + 2b α + 6c

= 2a α²– 3 a α² = – a α² from (i)

f(β) = 2aβ² + 2b β + 6c

= 2a β² + 3 a β² = 5a β²

f(α) f(β) = (- a α²) (5a β²)

= – 5 a² α² β² < 0

2ax² + 2bx + 6c = 0 has a root γ lying between α and β .

Q: If each pair of the following three equations x² + p₁x + q₁ = 0 , x² + p₂x + q₂ = 0 and x² + p₃ x + q₃ = 0 has exactly one root common, then prove that (p₁ + p₂ + p₃)2 = 4(p₁ p₂ + p₂ p₃ + p₃p₁ – q₁ – q₂ – q₃)

Solution : Let (α , β ) , (β ,γ) & (γ , α) be the roots of equation  .

⇒ α + β = – p₁  ,  α β = q₁

β + γ = – p₂ ,     β γ = q₂

γ + α = – p₃ ,     γ  α = q₃

R.H.S. = 4(p₁ p₂ + p₂ p₃ + p₃ p₁ – q₁ – q₂ – q₃)

= 4[(α + β) (β + γ) + (β + γ) (γ + α) + (γ + α) (α + β) – α β – β γ – γ  α ]

= 4[ α² + β² + γ² + 2(α β + β γ + γ  α) ]

= 4(α + β + γ )²

L.H.S. = (p₁ + p₂ + p₃)²

= ( – (α + β + β + γ + γ + α))²

= 4(α² + β² + γ²)

Q: If ax² + 2b x + c = 0 and a₁ x² + 2 b₁ x + c₁ = 0 have one and only one root in common, prove that b² – ac and b₁² – a₁ c₁ must be perfect square.

Solution : Since coefficients are rational. And roots cannot be complex or irrational because they occurs in pair.

Hence the roots must be real and rational. Which implies D must be perfect square of a rational number.

⇒ b² – ac as well as b₁² – a₁ c₁ is a perfect square of some rational number.

Q: For what integral values of a, the equation x² – x(1 – a) – (a + 2) = 0 has integral roots. Find the roots .

Solution: x²-x(1-a) – (a+2) – 0

⇒ (x – 1) (x + a) = 2

Since x and a are integers

Since both the factors in L.H.S. has to be integers

Possibilities are, x 1 = 1 and x + a = 2

⇒ x = 2 , a = 0

x – 1 = 2 and x + a = 1

⇒ x = 3 , a = – 2

x – 1 = -1 and x + a = – 2

⇒ x = 0 , a = – 2

x – 1 = -2 and x + a = – 1

⇒ x = – 1 , a = 0

Hence Possible integral values of a are – 2 , 0

For a = – 2 roots are 0, 3 and for a = 0 roots are –1 , 2.

Q: Show that the value of $\frac{tanx}{tan3x} $ , whenever defined, never lies between 1/3 and 3.

Solution : Let $\displaystyle y = \frac{tanx}{tan3x} $

$\displaystyle y = \frac{tanx}{\frac{3tanx – tan^3 x}{1-3 tan^2 x}} $

$\displaystyle y = \frac{tanx (1-3 tan^2 x)}{3tanx – tan^3 x} $

$\displaystyle y = \frac{(1-3 tan^2 x)}{3 – tan^2 x} $ ; as tanx ≠ 0

$\displaystyle 3 y – y tan^2 x = 1 – 3 tan^2 x $

$\displaystyle 3 y – 1 = (y – 3)tan^2 x $

$\displaystyle tan^2 x = \frac{3 y -1}{y -3}$

$\displaystyle tan^2 x = \frac{3 (y – 1/3)}{y -3}$

Since tan²x is positive, either y > 3 or y < 1/3. If y lies between 1/3 and 3 i.e. if 1/3 < y < 3, tan²x is negative, which is not possible.

Hence , $\displaystyle \frac{tanx}{tan3x}$ never lies between 1/3 and 3

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.