If alpha and bita are the zeroes of the polynomial 2y^2+7y+5 write the value of alpha + bita + alpha × bita
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2y^2+7y+5=2y^2+2y+5y+5 {factorising}
=2y(y+1)+5(y+1)
=(2y+5)(y+1)
therefore y=-5/2 and -1
therefore alpha=-5/2 and bita =-1
now alpha+bita+alphabita=-5/2+1+-5/2x1
=-5/2+1-5/2
=2(-5/2)+1
=-5+1
=-4
THEREFORE ANSWER IS -4
=2y(y+1)+5(y+1)
=(2y+5)(y+1)
therefore y=-5/2 and -1
therefore alpha=-5/2 and bita =-1
now alpha+bita+alphabita=-5/2+1+-5/2x1
=-5/2+1-5/2
=2(-5/2)+1
=-5+1
=-4
THEREFORE ANSWER IS -4
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