1(3)The given equation is
ax2+bx+c=0
its roots are -b+√(b2-4ac)2a and -b-√(b2-4ac)2a
The roots of the equation that we have to find are:
-b+√(b2-4ac)a and -b-√(b2-4ac)a
The required equation is x2-(sum of roots)x+product of roots=0
x2+2bax+4c=0
302. x= 2 + 21/3 + 22/3
x3- 6x2 + 6x = (x3- 6x2 + 12x - 8) + 8 - 6x
= (x - 2)3 - 6x + 8
= (21/3 + 22/3)3 - 6(2 + 21/3 + 22/3) + 8
= 2 + 22 + 6(21/3 + 22/3) - 6(2 + 21/3 + 22/3) + 8
= 2
11.Use cosine rule...
then:
-1<cos ∂<1
36For the first one... use the fact that,
the sum of two sides of a triangle is always greater than the third side.
For the second one... lemme do this....
x = 2 + 21/3 + 22/3 = 21/3 (1 + 21/3 + 22/3)
=> x3 = 2 (x - 1)3
=> x3 = 2 (x3 - 1 - 3x2 + 3x)
=> x3 = 2x3 - 6x2 + 6x - 2
=> x3 - 6x2 + 6x = 2 Ans...
For the third one...
Use this,
If the roots of the first eqns are a1,b1 then the roots of the second eqn.. will be (a1 + 2) and (b1 + 2)
and, quadratic equation with the roots m and n is given by
x2 - (m + n)x + mn done... !!