Question

Find the value of ‘k’ so that the zeros of the quadratic polynomial 3x2 – kx + 14 are in the ratio 7: 6.

Answer 1

hope you have understood .

Answer 2

Answer:

Let p(x)=3x2 -kx+14  

Let αα   and  ββ  are zeros of the given quadratic polynomial p(x).

Then A.T.Q.

α:β=7:6α:β=7:6  

So α=7mα=7m      and    β=6mβ=6m    where m is a real number.

On comparing 3x2 -kx+14 with ax2+bx+c we get  

a=3         b= -k      c=14

Now by using relationship between coefficients and zeros of a quadratic polynomial we get  

α+βα+β  = −ba−ba                and  α×βα×β  =  caca

α+βα+β   =  −(−k)3−(−k)3                  α×βα×β  = 143143

7m+6m= k3k3                             7m×6m= 143143

    13m =  k3k3                                      m2= 143×7×6143×7×6  

     3×13m = k                                    m2 = 1919  

         39m=k    ----(1)                        m2-1919  =0 ----(2)

From eq.(2)  we have

m2 −(13)2−(13)2   = 0

(m+1313)(m-1313)=0                   [BY USING IDENTITY (a2  - b2 )= (a+b)(a-b)]

m+1313 =0       or    m-1313  =0

     m= −13−13      or       m= 1313

From eq.(1)    k=39m

So,  If m= −13−13  then   k= 39m=39×(−13)(−13)  = -13        

       If m=1313        then   k= 39m=39×1313   =  13

Therefore there are two values of k=13  and -13