# 1.solve the quad eqn 2x² + kx - k²=0 for x. 2.if -5 is a root of the quad eqn 2x² + px -15=0 and the quad eqn p(x² + x) +k =0 has equal roots, find the value of k. 3.if the eqn (1 + m²)x² + 2mcx + (c² - a² ) =0 has equal roots, then prove that c² = a²(1 + m²). 4.Determine k so that (k² + 4k +8) , (2k² +3k +6) and (3k² +4k +4) are in A.P.

2
by simi

2016-01-15T21:30:38+05:30
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(1)2x²+kx-k²=0
2x²+2kx-kx-k²=0
2x(x+k)-k(x+k)=0
(x+k)(2x-k)=0
x=-k or x=k/2

(2)2x²-px-15=0
putting x=-5,we get
2(-5)²-p(-5)-15=0
50+5p-15=0
5p=-35
p=-7
Now putting p=-7 in other equation,we get
p(x²+x)+k=0
-7x²-7x+k=0
Roots are equal if D=0
b²-4ac=0
(-7)²-(7)(k)=0
49=7k
k=7

(3)(1+m²)x²+2mcx+(c²-a²)=0
Roots are equal if D=0
b²-4ac=0
(2mc)²-4(1+m²)(c²-a²)=0
4m²c²-4(c²-a²+m²c²-m²a²)=0
4m²c²-4c²+4a²-4m²c²+4m²a²=0
4m²c²-4m²c²-4c²=-4a²(1-m²)
-4c²=-4a²(1-m²)
Dividing both sides by -4,we get
c²=a²(1-m²)
Hence proved

(4)If the given terms are in AP then
(2k²+3k+6)-(k²+4k+8)=(3k²+4k+4)-(2k²+3k+6)
k²-k-2=k²+k-2
k+k=2-2
2k=0
k=0
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hey,wot about the 4 in the 12 th step? 4th qstn?
sry..not 4..2nd
2016-01-16T09:43:34+05:30
(1)2x²+kx-k²=0
2x²+2kx-kx-k²=0
2x(x+k)-k(x+k)=0
(x+k)(2x-k)=0
x=-k or x=k/2

(2)2x²-px-15=0
putting x=-5,we get
2(-5)²-p(-5)-15=0
50+5p-15=0
5p=-35
p=-7
substitute p=-7 in other equation,we get
p(x²+x)+k=0
-7x²-7x+k=0
Roots are equal if D=0
b²-4ac=0
(-7)²-(7)(k)=0
49=7k
k=7

(3)(1+m²)x²+2mcx+(c²-a²)=0
Roots are equal if D=0
b²-4ac=0
(2mc)²-4(1+m²)(c²-a²)=0
4m²c²-4(c²-a²+m²c²-m²a²)=0
4m²c²-4c²+4a²-4m²c²+4m²a²=0
4m²c²-4m²c²-4c²=-4a²(1-m²)
-4c²=-4a²(1-m²)
Dividing both sides by -4,we get
c²=a²(1-m²)
Hence proved

(4)If the given terms are in AP then
(2k²+3k+6)-(k²+4k+8)=(3k²+4k+4)-(2k²+3k+6)
k²-k-2=k²+k-2
k+k=2-2
2k=0
k=0