Question

a) If alfa ,betta ,gama are the zeroes of the Polynomial
ax3+bx2 +cx+d, then the value of the 1/alfa +1/bett +1/gama is​

Answer 1

EXPLANATION.

α, β, γ are the zeroes of the polynomial.

⇒ ax³ + bx² + cx + d.

As we know that,

Sum of zeroes of the cubic polynomial.

⇒ α + β + γ = -b/a.

Products of the zeroes of the cubic polynomial two at a time.

⇒ αβ + βγ + γα = c/a.

Products of the zeroes of the cubic polynomial.

⇒ αβγ = -d/a.

To find :

⇒ 1/α + 1/β + 1/γ.

⇒ βγ + αγ + αβ/αβγ.

Put the values in the equation, we get.

⇒ (c/a)/(-d/a).

⇒ c/a x a/-d. = -c/d.

⇒ 1/α + 1/β + 1/γ = -c/d.

                                                                                                                     

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

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★ Step by step explanation :

• Here, we have a cubic polynomial (ax³ + bx² + cx + d) whose zeroes are α,β,γ. We had to find out the value of 1/α + 1/β + 1/γ.

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~Sum of zeroes (α+β+γ) = -b/a

~Product of zeroes two at a time (αβ + βγ + γα) = c/a

~Product of zeroes (αβγ) = -d/a

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★ Finding value of 1/α + 1/β + 1/γ :

➲ㅤㅤㅤαβ + βγ+ γα/αβγ

★ Putting all known values :

➲ㅤㅤㅤc/a ÷ -d/a

➲ㅤㅤㅤc/a × a/-d

➲ㅤㅤㅤc/-d

•°• Value of 1/α + 1/β + 1/γ = c/-d

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★ More to know :

• Formula for finding roots of a quadratic equation = (-b±(√b²-4ac)/(2a).

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