Question

if the zeros of the polynomial p(x) = x³-9x²+18 are a-b,a,a+b, find a and b.​

Answer 1

Solution :-

We have the polynomial :-

x³ - 9x² + 18

As we know third degree polynomial can be written as

Ax³ + Bx² + Cx + D

Where

$\dfrac{-B}{A} = \alpha + \beta + \gamma$

= Sum of roots

$\dfrac{C}{A} = \alpha\beta + \beta\gamma + \gamma\alpha$

= Sum of Product of roots taken two at a time

$\dfrac{-D}{A} = \alpha\beta\gamma$

= Product of roots

So as in our polynomial :-

x³ - 9x² + 18

A = 1

B = -9

C = 0

D = 18

So as A = 1 , hence

Sum of roots = -B = α + β + γ

α + β + γ = -(-9)

α + β + γ = 9

Also it's Given

α = a - b

β = a

γ = a + b

→ α + β + γ = 9

→ ( a - b) + a + (a + b) = 9

→3a = 9

→ a = 3

Now from as product of roots (as A = 1)

= -D = αβγ

αβγ = -(18)

αβγ = -18

Now as

α = a - b

β = a

γ = a + b

→ (a -b)(a)(a + b) = -18

Also from above a = 3

→(3 - b)(3)(3 + b) = -18

→ 3( 3² - b²) = -18

→ 9 - b² = -6

→ -b² = -15

→ b² = 15

$b = \pm\sqrt{15}$

So Value of a and b

$\boxed{\begin{minipage}{4cm}a = 3 \\\\b = \pm\sqrt{15} \end{minipage}}$

Answer 2

Answer: a = 3  ; b =  ±√15.

Step-by-step explanation:

Given polynomial,

P(x): x³ - 9x² + 18

Comparing the following equation with standard form of 3rd degree equation Ax³ + Bx + C + D,

We have,

A = 1

B = -9

C = 0  

D = 18

Given that zeroes of the polynomial are (a - b) , a and (a + b)

We know that, for 3rd degree of equation,

Sum of roots  = -B/A

(a - b) + a + (a + b) = -(-9/1)

3a = 9

a = 9/3

a = 3 ,

Again,

Product of zeroes = -D/A

(a - b) × a × (a + b) =  -18/1

(3 - b) × 3 × (3 + b) = -18

(3 - b) × (3 + b) = -18/3

(3 - b)(3 + b) = -6

3² - b² = -6

9 - b² = -6

-b² = - 6 - 9

-b² = -15

b² = 15

b = ±√15.

Hence,

a = 3

b =  ±√15.