Find the value(s) of p for which the equation, (x+2)(x+p)=−1, has equal roots
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If the equation has equal roots, then the Discriminant will be 0.
Equation = ( x + 2 ) ( x + p ) = - 1
=> x^2 + px +2x + 2p = - 1
=> x^2 + ( P + 2 ) x + ( 2p +1 ) = 0
D = b^2 - 4ac
0 = ( p +2 )^2 - 4 × 1 × ( 2p + 1 )
0 = ( p^2 + 4 + 4p ) - ( 8p + 4 )
0 = p^2 + 4 + 4 p - 8p - 4
0 = p^2 - 4 p
0 = p ( p - 4 )
0 = p - 4
p = 4.
=> Value of p required is 4.
Equation = ( x + 2 ) ( x + p ) = - 1
=> x^2 + px +2x + 2p = - 1
=> x^2 + ( P + 2 ) x + ( 2p +1 ) = 0
D = b^2 - 4ac
0 = ( p +2 )^2 - 4 × 1 × ( 2p + 1 )
0 = ( p^2 + 4 + 4p ) - ( 8p + 4 )
0 = p^2 + 4 + 4 p - 8p - 4
0 = p^2 - 4 p
0 = p ( p - 4 )
0 = p - 4
p = 4.
=> Value of p required is 4.
John170912:
Why not p=0 ???
But p = 4, then how it can be 0.
P(P-4)=0
Which implies p=0 and p=4
Oh.. ya.. here p = 0. and p = 4.
Ya..btw thanxs
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