If p,q,r are zeroes of the polynomial p(x) = 9x^3-3x^2-7x+1 then find p^-1+q^-1+r^-1
Answers
Step-by-step explanation:
1, 7.246, - 1.25
Step-by-step explanation:
The cubic polynomial is 9x³ - 3x² - 7x + 1 and p, q, and r are the zeros of the equation.
Therefore, if 9x³ - 3x² - 7x + 1 = 0, then x = p, x = q and x = r will be the roots of the equation.
Using vanishing method x = 1 makes the equation valid.
Hence, (x - 1) will be a factor of the expression.
So, 9x³ - 3x² - 7x + 1 = 0
⇒ 9x³ - 9x² + 6x² - 6x - x + 1 = 0
⇒ 9x²(x - 1) + 6x(x - 1) -1(x - 1) = 0
⇒ (x - 1)(9x² + 6x - 1) = 0
Hence, (x - 1) = 0 or (9x² + 6x - 1) = 0
Now, x = 1 or (9x² + 6x - 1) = 0
For, (9x² + 6x - 1) = 0
⇒ x = \frac{-6 \pm \sqrt{6^{2} - 4(9)(-1) } }{2(9)}x=
2(9)
−6±
6
2
−4(9)(−1)
{Applying the quadratic formula}
⇒ x = 0.138 and x = - 0.8
Therefore, p = 1, q = 0.138 and r = -0.8
Hence, p^{-1} = (1)^{-1} = 1p
−1
=(1)
−1
=1
q^{-1} = (0.138)^{-1} = 7.246q
−1
=(0.138)
−1
=7.246
and r^{-1} = (- 0.8)^{-1} = - 1.25r
−1
=(−0.8)
−1
=−1.25 (Answer)
Answer:
7
Step-by-step explanation:
pq+qr+pr =c/a = -7/9 ---- (1)
pqr = -d/a = -1/9 ----- (2)
1/p+1/q+1/r = (pq+qr+pr)/pqr
= (-7/9) /(-1/9) ----- [from 1 & 2]
= 7