find the value of p for which the quadratic
(2P+ 1) x² - (7 P + 2) x + (7P-3) = 0 has equal
roots. Also find these words.
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Answer:
a= 2p +1 , b= 7p +2 , c= 7p -3
D=0
D= under root b^2 - 4ac
squaring both side
0^2= (7p+2)^2 - 4 x 2p+1 x 7p-3
0. = (49p^2 + 4 + 2 x 7p x 2 ) - (56p^2 + 4p -12)
0. = 49p^2 +4 + 28p -56p^2 -4p +12
0. = 7p^2 + 24p +16
0= 7p^2 +28p - 4p +16
0= 7p( p+4) -4(p+4)
0= ( 7p-4) ( p+4)
p=7/4. p=-4
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