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Question

Let a,b (b>a) are the roots of the quadratic equation (k+1)x2(20k+14)x+91k+40=0; where k>0, then which among the following option(s) is/are correct for the roots

A
a(4,7)
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B
b(4,7)
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C
a(7,10)
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D
b(10,13)
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Solution

The correct options are
A a(4,7)
D b(10,13)
(k+1)x2(20k+14)x+91k+40=0;k>0
The equation can be rewritten as
x220k+14k+1x+91k+40k+1=0
Let f(x)=x220k+14k+1x+91k+40k+1
f(x)=x2(20(k+1)6k+1)x+91(k+1)51k+1f(x)=(x220x+91)+(6x51k+1)f(x)=(x13)(x7)+(6x51k+1)

Now,
f(4)=(9)(3)27k+1f(4)=27kk+1>0 (k>0)f(7)=9k+1<0f(10)=(3)(3)+9k+1f(10)9kk+1<0f(13)=27k+1>0
So, the graph of the polynomial


The equation has two roots- one in (4,7) and another in (10,13)

Hence, as b>a so a(4,7), b(10,13)

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