if alpha beta are zeroes of polynomial x²-4x-12 then find 1/alpha + 1/beta -2ab
Answers
EXPLANATION.
α and β are zeroes of the polynomial.
⇒ x² - 4x - 12 = 0.
As we know that,
Sum of the zeroes of the quadratic equation.
⇒ α + β = -b/a.
⇒ α + β = -(-4)/1 = 4.
Products of the zeroes of the quadratic equation.
⇒ αβ = c/a.
⇒ αβ = (-12)/1 = -12.
To find : 1/α + 1/β - 2αβ.
Taking L.C.M. in this equation, we get.
⇒ β + α - 2αβ(αβ)/αβ.
⇒ α + β - 2(αβ)²/αβ.
Put the values of the equation, we get.
⇒ 4 - 2(-12)²/-12.
⇒ 4 - 2(144)/-12.
⇒ 4 - 288/-12.
⇒ -284/-12.
⇒ 71/3.
⇒ 1/α + 1/β - 2αβ = 71/3.
MORE INFORMATION.
Conjugate roots.
If D < 0.
One roots = α + iβ.
Other roots = α - iβ.
If D > 0.
One roots = α + √β.
Other roots = α - √β.
α and β are the zeroes of the polynomial x² + 4x + 3.
So,
Sum of zeroes :
α + β = -b/a
⇒ α + β = -4
Product of zeroes:
αβ = c/a
⇒ αβ = 3
Now,
★Sum of zeroes :
1 + β/α + 1 + α/β
⇒ αβ + β² + αβ + α² / αβ
⇒ α² + β² + 2αβ / αβ
We know that,
α² + β² + 2αβ = ( α + β )²
⇒ ( α + β )² / αβ
Putting the values
⇒ ( - 4 )² / 3
⇒ 16 / 3
★ Product of zeroes:
1 + β/α × 1 + α/β
⇒ 1 + α/β + β/α + αβ/αβ
⇒ 2αβ + α² + β² / αβ
⇒ ( α + β )² / αβ
Putting the values
⇒ ( - 4 )² / 3
⇒ 16 / 3
But required polynomial is:
x² - ( Sum of the zeroes ) x + Product the zeroes
⇒ x² - 16/3 x + 16/3
If k = 3
⇒ 3 ( x² - 16/3 x + 16/3 )
⇒ 3x² - 16x + 16