1.

find the zeroes of quadratic polynomial 6x²-7x-3and verify the relationship between the zeroes and their coefficient.​

Answer»

<P>Answer->

Let p(x)=6x2-7x-3

Now,

Let p(x)=0

=>6x2-7x-3=0

Doing MIDDLE TERM SPLITTING,we get,

6x2-9x+2x-3=0

=>3x(2x-3)+1(2x-3)=0

=>(2x-3)(3x+1)=0

Therefore,

(2x-3)=0 ; (3x+1)=0

=>2x-3=0 ; 3x+1=0

=>2x=3 ; 3x=-1

=>x=3/2 ; x=-1/3

So, The two zeroes of p(x) are 3/2 and -1/3

Verification->

Sum of the zeroes=3/2+(-1/3)

= 9/6+(-2/6) {Taking L.C.M of 3 and 2}

=7/6

=-coefficient of x/coefficent of x2

And,

Product of the zeroes=3/2*(-1/3)

=-3/6

=CONSTANT term/coefficient of x2

Hence verified.


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