Quadratic Equation -(2nd Part) --- By K. Bhanumoorthy

                                                       Quadratic Equation ------continued.

“In an effective class room, students should, not only know what they are doing, they should also know why and how.”----------------Harry Wong.

Some problems reducing to Quadratic Equation form:

1. Find the values of X for which the given equation is true.

4X – 5 (2(X+3)) + 256 = 0.

Explanation:- Let us put 2X = Y, then for the given statement }

 The equation is (2X)2 - 5 (2X. 23) + 256 = 0, reduced to the equation 

Y2 - 40Y + 256 = 0, By factorizing we get (Y –32) (Y—8) = 0; Y = 32, Y = 8,

i.e.   2X = 32, this is true for X = 5, similarly 2X =8, this is true for X = 3.

 True values   of X are 3 and 5. Answers.

2. Given two quadratic equations. One of the roots is common between them. Find the integral values of p and q. The roots are given to be, that they are in A.P

Equations are (i) X2 –3X + p=0, (ii) X2 –5 X + q=0 

Explanation: -Let α, β be two roots of equations, then sum of roots   α + β = 3   (i)  and product of roots 

   α x β  = p (ii) 

Let β be common root. So, let  β and γ be the roots of equation (ii).

Sum   β + γ = 5 (iii), and product β x γ = q  (iv)

Given roots are in A.P. So, If they are in AP, 2β = α +  γ  or β = (α + γ) / 2;

α + { (α + γ ) / 2}  = 3; 3α  + γ  =   6 (v)  ,by solving   from (i),(iii) and (v)

we get   α = 1, β = 2, and  γ = 3, hence p = 2 and q = 6. 

3. Given two equations X2 - 7X + a = 0 and X2 - 13X + 4a = 0. One of the roots is common. Find the values of a.

Explanation:- Let the common root be α. It must satisfy both the given equations.

Hence  α2 – 7X + a = 0  (i)  and    α2 –13α + 4a  = 0 (ii), 

Subtracting (i) from (ii) we  get α = a/2, let us put the value of α . in equation (i), we get (a/2)2 – 7(a/2) + a = 0, after 

taking LCM and simplification, we get a2 – 14a + 4a = 0, a2 – 10a = 0,

a(a-10) = 0, either a = 0 or a - 10 = 0, ie. a= 0, a =10. Values of a: 0 and 10

4. Given, α, β,  be the roots of X2 -2X +p = 0.  γ , δ be the roots of X2 – 8X +q = 0

If α, β, γ, δ are in G.P. Find the integral values of p and q.

 Explanation:- Let r be the common ratio for the given G. P, then, α, β, γ, δ  becomes α, αr, αr2, αr3. We can form the sum and product of the roots 

 α + α r = α (1+r) = 2  (i),   

α x αr = p  (ii).

γ +  δ = 8 ,   γ.  δ = q 

αr2+ αr3 = 8, αr2(1 +r) = 8 (iii); αr2(1 +r) ÷ α (1+r) = 8 / 2; we get r2 = 4, hence r = +2 or -2.

There are two cases, If r = 2 then , putting in equation (i) we get  3 α = 2, so α = 2 / 3.

Further the value of α β= α2r = (4 /9) x 2 = 8 /9 =p, here p is not an integer, hence r = 2 does not satisfy for the conditions p and q to be integers in the question.

Let us put r = -2.   α (1+r) = 2, α (1 – 2 ) = 2 , hence, α = -2

The value of αβ = α. αr = α2r = (-2) (-2). (-2) = -8; so, the value of p is -8

In the same way taking the second equation the product of

 γ  δ =   αr2, αr3 = ( -2)(-22).( -2) (-23 ) =   -128 ,which is equal to q .

Hence the integral values of p and q are -8 and -128, which satisfy the given conditions that the roots α, β, γ, δ are in GP   ie  the  answer is :-2, 4,-8, 16. 

5. If (n+1) n! = 90 x (n-1)! find n.

(n+1) n (n-1)! = 90 (n-1)!

Cancelling the common term, we get

n(n+1) = 90

n2 + n – 90 = 0

can be factorized as,

(n + 10) (n - 9) = 0

n = 9, or n= -10

Hence the answer is n = 9. -10 is neglected. (We use only natural numbers for factorials)