Two real numbers α β are such that α+β=3 and difference of α and β is 4 , then α β are the monts of the quadratic equation:

Question

Two real numbers α β are such that α+β=3 and difference of α and β is 4 , then α β are the monts of the quadratic equation: (A) 4 x2-12 x-7=0 (B) 4 x2-12 x+7=0 (C) 4 x2-12 x+25=0 (D) 4 x2+12 x+7=0

Tardigrade Admin · Accepted Answer

|α-β|=4 ⇒ (α-β)2=16 ⇒ (α+β)2-4 α β=16 ⇒ 9-4 α β=16 ⇒ α β=-7 / 4 ⇒ equation is x2-3 x-(7/4)=0

Q. Two real numbers $α & β$ are such that $α + β = 3$ and difference of $α$ and $β$ is 4 , then $α & β$ are the monts of the quadratic equation:

$4 x^{2} - 12 x - 7 = 0$

$4 x^{2} - 12 x + 7 = 0$

$4 x^{2} - 12 x + 25 = 0$

$4 x^{2} + 12 x + 7 = 0$

$∣ α - β ∣ = 4$
$\Rightarrow (α - β)^{2} = 16$
$\Rightarrow (α + β)^{2} - 4 α β = 16$
$\Rightarrow 9 - 4 α β = 16$
$\Rightarrow α β = - 7/4$
$\Rightarrow$ equation is $x^{2} - 3 x - \frac{7}{4} = 0$

Q. Two real numbers α&β are such that α+β=3 and difference of α and β is 4 , then α&β are the monts of the quadratic equation:

4x2−12x−7=0

4x2−12x+7=0

4x2−12x+25=0

4x2+12x+7=0