Question

answer these questions

Answer 1

20)
(x-1)^2+(x-2)^2+(x-3)^2=0
3x^2-12x+14=0
Discriminant=b^2-4ac
=(-12)^2-4(3)(14)
=144-168
=-24
=<0
Hence roots are imaginary.


21)
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
3x^2-2bx-2ax-2cx+ab+bc+ac=0
x^2-2x(b+a+c)+(ab+bc+ac)=0
Given that roots are equal.
Therefore,
Discriminant=0
b^2-4ac=0
{-2(a+b+c)}^2-4(3)(ab+bc+ac)=0
{4(a^2+b^2+c^2+2ab+2bc+2ac)}-4(3ab+3bc+3ac)=0
a^2+b^2+c^2+2ab+2bc+2ac-3ab-3ac-3bc=0
a^2+b^2+c^2-ab-bc-ac=0
a^2+b^2+c^2=ab+bc+ac
a×a+b×b+c×c=ab+bc+ac...........................1
comparing a,b,c on both sides we get,
a=b, b=c, c=a
By above we get,
a=b=c
we can also verify above answer by putting above equation in eq1 as answer is correct equation will get satisfied.



22)
i)2x^2+3x-1=0
Discriminant=b^2-4ac=17>0
Hence roots are real and distinct.

ii)x^2+x+1
Discriminant=b^2-4ac=-3<0
Hence roots are imaginary.

iii)4x^2+4x+1
Discriminant=b^2-4ac=16-16=0
Hence roots are real and equal.


23)
let p(x)=x^2+px-4=0
Given that,-4 is the zero of p(x)
Then we have,
p(-4)=0
(-4)^2+p(-4)-4=0
16-4p-4=0
p=3
also given that
x^2+px+k=0 have equal roots.
putting value of p the above polynomial becomes,
x^2+3x+k=0
Discriminant=0
b^2-4ac=0
9-4k=
k=9/4
Hence value of k=9/4.