LMNO is a rhombus, seg OP is perpendicular to side LM and
seg OQ
is perpendicular to side MN. Prove that
seg OP is congruent to seg OQ...
plz ans...need it urgent..
Answers
Answered by
1
Answer:
This implies that
x2+2ax=4x−4a−13
or
x2+2ax−4x+4a+13=0
or
x2+(2a−4)x+(4a+13)=0
Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.
Hence we get that
(2a−4)2=4⋅1⋅(4a+13)
or
4a2−16a+16=16a+52
or
4a2−32a−36=0
or
a2−8a−9=0
or
(a−9)(a+1)=0
So the values of a are −1 and 9.
Answered by
1
Answer:
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