Let roots of equation: ax 2+ bx + c =0 are (α-β) and (γ-δ) and roots of equation Ax 2+ Bx + C =0 are (α+δ) and (β+γ) then |( a / A )| equals [Note: D 1, D 2 represent discriminants of ax 2+ bx + c =0 and Ax 2+ Bx + C =0 respectively.

Question

Let roots of equation: ax 2+ bx + c =0 are (α-β) and (&gamma;-δ) and roots of equation Ax 2+ Bx + C =0 are (α+δ) and (β+&gamma;) then |( a / A )| equals [Note: D 1, D 2 represent discriminants of ax 2+ bx + c =0 and Ax 2+ Bx + C =0 respectively. (A) |(b/B)| (B) |( c / C )| (C) √|(D1/D2)| (D) √|(D2/D1)|

Tardigrade Admin · Accepted Answer

Difference of roots is same ⇒ S 12-4 P 1= S 22-4 P 2 ⇒ ( b 2/ a 2)-(4 c / a )=( B 2/ A 2)-(4 C / A ) ⇒ (( a / A ))2=( b 2-4 ac / B 2-4 AC )=( D 1/ D 2)

$∣ ∣ \frac{b}{B} ∣ ∣$

$∣ ∣ \frac{c}{C} ∣ ∣$

$∣ ∣ \frac{D _{1}}{D _{2}} ∣ ∣$

$∣ ∣ \frac{D _{2}}{D _{1}} ∣ ∣$

Difference of roots is same
$\Rightarrow S_{1}^{2} - 4 P_{1} = S_{2}^{2} - 4 P_{2}$
$\Rightarrow \frac{b ^{2}}{a ^{2}} - \frac{4 c}{a} = \frac{B ^{2}}{A ^{2}} - \frac{4 C}{A}$
$\Rightarrow (\frac{a}{A})^{2} = \frac{b ^{2} - 4 a c}{B ^{2} - 4 A C} = \frac{D _{1}}{D _{2}}$

Q. Let roots of equation : ax2+bx+c=0 are (α−β) and (γ−δ) and roots of equation Ax2+Bx+C=0 are (α+δ) and (β+γ) then ∣∣​Aa​∣∣​ equals [Note: D1​,D2​ represent discriminants of ax2+bx+c=0 and Ax2+Bx+C=0 respectively.

∣∣​Bb​∣∣​

∣∣​Cc​∣∣​

∣∣​D2​D1​​∣∣​​

∣∣​D1​D2​​∣∣​​

Difference of roots is same ⇒S12​−4P1​=S22​−4P2​ ⇒a2b2​−a4c​=A2B2​−A4C​ ⇒(Aa​)2=B2−4ACb2−4ac​=D2​D1​​

$∣ ∣ \frac{b}{B} ∣ ∣$

$∣ ∣ \frac{c}{C} ∣ ∣$

$∣ ∣ \frac{D _{1}}{D _{2}} ∣ ∣$

$∣ ∣ \frac{D _{2}}{D _{1}} ∣ ∣$

Difference of roots is same
$\Rightarrow S_{1}^{2} - 4 P_{1} = S_{2}^{2} - 4 P_{2}$
$\Rightarrow \frac{b ^{2}}{a ^{2}} - \frac{4 c}{a} = \frac{B ^{2}}{A ^{2}} - \frac{4 C}{A}$
$\Rightarrow (\frac{a}{A})^{2} = \frac{b ^{2} - 4 a c}{B ^{2} - 4 A C} = \frac{D _{1}}{D _{2}}$