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