Transformation of Roots: Algebraic Transformation
Trending Questions
- x2+2x+3=0
- x2+4x−4=0
- x2+4x+4=0
- x2−4x+4=0
If α, β are the roots of the equation 2x2−3x−5=0, then the equation whose roots are 5α , 5β is :
x2−3x−10=0
x2−3x+10=0
x2+3x+10=0
x2+3x−10=0
- always imaginary
- opposite in sign
- same in sign
- always real
- (2m−1)x2−6x+m=0
- 6x2−(2m−1)x−m=0
- mx2+(2m−1)x+6=0
- (2m−1)x2+6x+m=0
- 21x2+5x−1=0
- 147x2+35x−5=0
- 21x2+5x−3=0
- 147x2+35x−14=0
- (p3+q)x2−(p3+2q)x+(p3+q)=0
- (p3−q)x2−(5p3+2q)x+(p3−q)=0
- (p3+q)x2−(5p3−2q)x+(p3−q)=0
- (p3+q)x2−(p3−2q)x+(p3+q)=0
- aq=cr
- bp=aq
- aq=bp
- ap=rc
If α, β and γ are the roots of the equation x3+3x+2=0...(i) and (α−β)(α−γ), (β−γ)(β−α), (γ−α)(γ−β) are the roots of equation y3−9y2−216=0...(ii).
Relation to transform the eq.(i) to eq.(ii) is given by .
y=α2−1α
y=2α2−1α
y=2α2−2α
y=α2−2α
- (p3−q)x2−(p3+2q)x+(p3−q)=0
- (p3+q)x2−(p3+2q)x+(p3+q)=0
- (p3−q)x2−(p3−2q)x+(p3+q)=0
- (p3+q)x2−(p3−2q)x+(p3+q)=0
- x2−5x+4=0
- x2−3x−2=0
- x2−13x+4=0
- x2+5x+4=0
Let α, β be the values of m for which the equation (1+m)x2−2(1+3m)x+(1+8m) has equal roots. Find the equation whose roots are α+2 and β+2.
x2−5x+6=0
x2−3x=0
x2−7x+10=0
x2−4x+3=0
Find the equation whose roots are reciprocals of the roots of 2x2+5x+4=0 .
4x2+2x+5=0
5x2+4x+2=0
2x2+4x+5=0
4x2+5x+2=0
- x2−4x−4=0
- x2−4x+4=0
- x2+4x−4=0
- x2+4x+4=0
- x2−6x+25=0
- x2+10x+25=0
- x2−26x+25=0
- x2+6x+25=0
- None of the above
- a(√x)4+b(√x)2+c=0
- (a√x)2+(b√x)+c=0
- a(√x)2+b(√x)+c=0
- x4−5x2−4=0
- 4x4−5x2+1=0
- x4+5x2+4=0
- 4x4+5x2+1=0
- 0
- 2
- 1
- 3
If α, β and γ are the roots of the equation x3+2x2+3x+1=0 . Find the equation whose roots are α3, β3 and γ3.
x3−5x2+5x+2=0
x3−x2+15x+2=0
x3−7x2+12x+1=0
x3−2x2+10x+1=0
- x2−5x+4=0
- x2−13x+4=0
- x2−3x−2=0
- x2+5x+4=0
If α, β, γ and δ are the roots of x4+2x3+3x2+4x+5=0.
Find the equation whose roots are 1α, 1β, 1γ, 1δ.
x4+3x3+2x2+4x+5=0
5x4+4x3+3x2+2x+1=0
5x4−4x3+3x2−2x+1=0
x4−2x3+3x2−4x+5=0
If the roots of the equation ax2+bx+c=0 beα and β, , then the roots of the equation cx2+bx+a=0 are
1α, 1β
−α, −β
None of these
α, 1β
- 27x2+48x+2=0
- 27x2+46x+2=0
- 46x2+27x+2=0
- 27x2+2x+48=0
If α, β are the roots of the equation 2x2+5x+6=0, then find the equation whose roots are 1α and 1β.
5x2−6x+2=0
6x2+5x+2=0
6x2−5x+2=0
5x2+6x+2=0
- x4+17x2+16=0
- x4−17x2−16=0
- 16x4−17x2+1=0
- 16x4+17x2+16=0
- 4x2+7x+16=0
- 4x2−7x−16=0
- 4x2−7x+16=0
- 4x2+7x−16=0
- None of these
- acx2+(a+c)cx+(a+c)2=0
- abx2+(a+c)bx+(a+c)2=0
- acx2+(a+c)bx+(a+c)2=0
- a:b:c=1:2:−2
- a=−c
- b=−c
- a:b:c=2:1:−2