Are you watching this Quadratic Equation Closely?

Algebra Level 3

The quadratic equation 2 x 2 + 6 x + γ 2x^{2} + 6x + \gamma , where γ < 0 \gamma < 0 , has roots α \alpha and β \beta .

Then the value of α β + β α \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} is:

2 0 ≤ − 2 \leq -2 < − 2 < -2 -2 ≥ 2 \geq 2

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Sabhrant Sachan
May 29, 2016

From the quadratic equation 2 x 2 + 6 x + γ < 0 α + β = − 3 and α ⋅ β = γ 2 α β + β α ⟹ α 2 + β 2 α ⋅ β ⟹ ( α + β ) 2 α ⋅ β − 2 ⟹ 18 γ − 2 γ < 0 ⟹ γ 18 < 0 ⟹ 18 γ < 0 18 γ − 2 < − 2 \text{From the quadratic equation } 2x^2+6x+\gamma <0 \\ \alpha+\beta = -3 \text{ and } \alpha\cdot\beta=\dfrac{\gamma}{2} \\ \dfrac{\alpha}{\beta}+\dfrac{\beta}{\alpha} \implies \dfrac{\alpha^2+\beta^2}{\alpha\cdot\beta} \implies \dfrac{(\alpha+\beta)^2}{\alpha\cdot\beta}-2 \implies \dfrac{18}{\gamma}-2 \\ \gamma<0 \implies \dfrac{\gamma}{18}<0 \implies \dfrac{18}{\gamma}<0 \\ \boxed{\dfrac{18}{\gamma}-2<\color{#3D99F6}{-2}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...