α,β are the two roots of the equation 2x² + x + 1 = 0, find an equation whose roots are
α²/β , β²/α
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EXPLANATION.
α,β are the roots of the equation,
⇒ F(x) = 2x² + x + 1 = 0.
As we know that,
Sum of zeroes of quadratic equation,
⇒ α + β = -b/a.
⇒ α + β = -1/2.
Products of zeroes of quadratic equation,
⇒ αβ = c/a.
⇒ αβ = 1/2.
Equation whose roots are,
⇒ α²/β, β²/α.
Sum of zeroes of quadratic equation,
⇒ α²/β + β²/α.
⇒ α³ + β³/αβ.
As we know that,
Formula of :
⇒ α³ + β³ = (α + β)(α² - αβ + β²).
⇒ α² + β² = (α + β)² - 2αβ.
Using both formula in equation, we get.
⇒ (α + β)(α² - αβ + β²)/αβ.
⇒ (α + β)[ (α + β)² - 2αβ - αβ]/αβ.
⇒ (α + β)[(α + β)² - 3αβ]/αβ.
Put the values in the equation, we get.
⇒ (-1/2)[(-1/2)² - 3(1/2)]/1/2.
⇒ (-1/2)[1/4 - 3/2]/1/2.
⇒ (-1/2)[1 - 6/4]/1/2.
⇒ (-1/2)[-5/4]/1/2.
⇒ 5/8/1/2.
⇒ 5/8 X 2/1.
⇒ 5/4.
Products of zeroes of equation,
⇒ α²/β, β²/α.
⇒ α²/β X β²/α.
⇒ αβ.
⇒ 1/2.
Equation of quadratic polynomial,
⇒ x² - (α + β)x + αβ.
Put the values in the equation, we get.
⇒ x² - (5/4)x + 1/2 = 0.
⇒ 4x² - 5x + 2 = 0.
MORE INFORMATION.
Quadratic expression.
A polynomial of degree two of the form ax² + bx + c (a ≠ 0) is called a quadratic expression in x.
The quadratic equation.
ax² + bx + c = 0 (a ≠ 0) has two roots, given by.
α = -b + √D/2a.
β = -b - √D/2a.
D = Discriminant.
D = b² - 4ac.
◉GIVEN:-
- α and β are the roots
- equation -- 2x^2 + x + 2=0
◉TO FIND:-
◉ SOLUTION:-
our standard equation is
ax^2 +bx+c =0
Let's compare--
- a = 2
- b = 1
- c = 1
roots of the equation 2x^2 + x +1=0 are α and β
Now,
- α+β = -b/a = -1/2
- αβ = c/a = 1/2
So,
~~ α^2/β + β^2/α =
~~α^3+ β^3/ αβ=
~~ (α+β)^3- 3αβ(α+β)/αβ=
putting the values:-
~~ (-1/2)^3 -3 × 1/2(-1/2)/1/2=
~~( -1/8 + 3/4) × 2=
~~ 2(-1+6/8)=
~~ 5/4
and,
~~ α^2/β × β^2/α=
~~ α^2 β^2/αβ=
~~ αβ =
~~ 1/2
Now, the equation is,
x^2 -(Addition of two roots)X + (products of roots)=0
✔️ x^2 - 5/4x + 1/2 = 0
✔️ 4x^2 -5x +2=0
So, the required equation is
4x^2 -5x+2=0