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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.

...................... (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 ...................................

(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