help.. 2 questions (1 Viewer)

[muttiah]

Show that the roots of one of the eqns

8ax^2(2x-1)+b^2=0, 4a^2x^2 +b^2(4x+1) = 0 must b unreal..

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Show that the eqn (x+1)(x-4)=mx has 2 distinct roots for all real values of the constant m

 

[Trebla]

For the second one:
(x + 1)(x – 4) = mx
x² - 3x - 4 = mx
x² - 3x - mx - 4 = 0
x² + x(- 3 - m) - 4 = 0
Δ = (- 3 - m)² + 16
Δ = (m + 3)² + 16
Since (m + 3)² ≥ 0 for all real m, therefore (m + 3)² + 16 > 0
Therefore Δ > 0, hence equation has 2 distinct roots

 

[jyu]

Show that the roots of one of the eqns

8ax^2(2x-1)+b^2=0, 4a^2x^2 +b^2(4x+1) = 0 must b unreal..

Please type your question carefully.

Is the first equation 8(a^2)x(2x-1)+b^2=0 ?

:wave:

 

[jyu]

If the first equation has real roots, its discriminant >= 0,
.: a^2 - b^2 >= 0 , assuming a =/= b (note: I think this is given),
then a^2 - b^2 > 0 and the discriminant of the second equation < 0,
.: the second equation has unreal roots.

If the first equation has unreal roots, its discriminant < 0,
then a^2 - b^2 < 0 and the discriminant of the second equation > 0,
.: the second equation has real roots.

:wave:

 

[pyrodude1031]

FOr the second question ,

analysing graphically.. considering lhs and rhs ..

the lhs = (x+1)(x-4) which is a parabola cutting x=-1,4

and the rhs =mx

the equality between simply means that the x values are the intersections between the parabola and a straight line crossing the origin with gradient m ..

by inspection, any straightline of any real valued gradient m CROSSING the origin must cut the parabla twice... HEnce result follows.