Question

if p q r are the zeros of the polynomial 9x^3-3x^2-7x+1 then find p^-1, q^-1, r^-1

Answer 1

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

⇒ $x = \frac{-6 \pm \sqrt{6^{2} - 4(9)(-1) } }{2(9)}$ {Applying the quadratic formula}

⇒ x = 0.138 and x = - 0.8

Therefore, p = 1, q = 0.138 and r = -0.8

Hence, $p^{-1} = (1)^{-1} = 1$

$q^{-1} = (0.138)^{-1} = 7.246$

and $r^{-1} = (- 0.8)^{-1} = - 1.25$ (Answer)

Answer 2

answer is 7

the attachment is below

hope it helps