If c≠0 and the equation p2x=a(x+c)+b(x-c) has 2 equal roots, then p can be
a+b
a-b
(a±b)2
None of these
Explantion for the correct options:
Find the value of p:
Given, p2x=ax+c+bx-c
⇒ p2x=ax-c+bx+cx+cx-c
⇒ p2x=ax-ac+bx+bcx2-c2
⇒ px2-c2=2xax-ac+bx+bc
⇒ px2-pc2=2ax2-2acx+2bx2+2bcx
⇒px2-pc2-2ax2+2acx-2bx2-2bcx=0
⇒px2-2ax2-2bx2+2acx-2bcx-pc2=0
⇒ p-2a-2bx2+2ac-bcx-pc2=0
For equal roots, 4ac-bc2+4pc2p-2a-2b=0
a-b2+p2-2pa+b=0
p-a+b2=a+b2-a-b2=4ab
∴p=a+b±2ab=(a±b)2
Hence, option(C) is correct.
If the equation 2x2 + 2(c – 20) x + (c – 20) = 0 has equal roots then, the value of c is
If a,b,c∈R and a≠0, c<0, and if the quadratic equation ax2+bx+c=0 has imaginary roots, then a+b+c is