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Find the zeroes of the quadratic polynomial x2 + 5x + 6 and verify the relationship between the zeroes and the coefficients.

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Let P(x)=x2+5x+6=0 x2+3x+2x+6=0 (x+3)(x+2)=0 x=-3,-2 If ax2+bx+c=0 is a general form of polynomial with degree 2 Sum of zeroes is -b/a Product of zeroes is c/a From the above polynomial Sum of zeroes -3-2=-5=-5/1=-b/a Product of zeroes=-3×-2=6=6/1=c/a
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-3 & -2
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Let p (x) = x2+ 5x + 6 Now x2 + 5x + 6 = 0 x2 + 3x + 2x + 6 = 0 x ( x + 3 ) + 2 ( x + 3 ) = 0 X= - 3 and - 2 Then put value of x you get result
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x=-2,x=-3;-b\a=-5\1=-5;c\a=6\1=6
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Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha...
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Let p(x)= x2+5x+6 Now putting p (x)=0 X2+5x+6=0 X2+3x+2x+6=0 X (x+3)+2 (x+3)=0 (X+3)(x+2)=0 (X+3)=0,(x+2)=0 X=-3,x=-2 Therefor -3 and -2 are zeros of given polynomial p (x)and names alpha and beta Now we have to verify the cofficients of given polynomial Comapre p (x)=ax2+bx+c ;a=1,b=5,c=6 Now alpha +beta= -b/a Alpha*beta=c/a For the verification of cofficients take both the scenario L.H.S alpha+beta and R.H.S -b/a -3 + - 2 -5/1 -5 -5 Hence L.H.S=R.H.S Now Again check for L.H.S alpha*beta R.H.S c/a -3 * -2 6/1 6 6 Hence L.H.S = R.H.S So the relationship between zeros and cofficients of given polynomial is verified. read less
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