Question

If the zeroes of the polynomial x3 – 3x2 + x + 1 are a-b, a, a + b, find a and b.
❌ no spam pleach ♡​

Answer 1

ᴄᴏʀʀᴇᴄᴛ qᴜᴇꜱᴛɪᴏɴ:

• The zeroes of the polynomial
• x³ – 3x² + x + 1 are— a-b, a, a + b

ᴛᴏ ꜰɪɴᴅ:

• A and B

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

★For a Cubic Polynomial:

$~~~~~~~~~$ p(x) = ax³ + bx² + cx + d

★With zeroes $\alpha$ , $\beta$ and $\gamma$

We have,

$\alpha$ + $\beta$ + $\gamma$ = $\frac{ - b}{a}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

$\alpha \beta \gamma = \frac{ - d}{a}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

NOW-

$~~~~~~~~~$ p(x) = x³ - 3x² + x + 1

★Comparing with p(x) = Ax³ + Bx² + Cx + D

A = 1

B = -3

C = 1

D = 1

Zeroes are—

$\alpha = a - b$

$\beta = a$

$\gamma = ab$

★Sum Of Zeroes—

$~~~~~~~~~$ $\implies$ Sum of zeroes = $\frac{ - b}{a}$

$~~~~~~~~~$ $\implies$ $\alpha + \beta + \gamma = \frac{ - b}{a}$

$~~~~~~~~~$ $\implies$ a - b + a + a + b = 3

$~~~~~~~~~$ $\implies$ 3a = 3

$~~~~~~~~~$ $\implies$ a = 1

★Sum Of Product Zeroes—

Sum of product zeroes = $\frac{c}{a}$

$\implies$ $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$

⟶ (a-b)a + a(a+b) + (a+b) (a-b) = 1

⟶ a² - ba + a² + ab + a² - b² = 1

⟶ a² + a² + a² - b² = 1

⟶ 3a² - b² = 1

PUTTING A=1

⟶ 3(1)² - b² = 1

⟶ 3 - b² = 1

⟶ 3 - a = b²

⟶ b² = 2

⟶ b = $± \sqrt{2}$

Therefore,

• a = 1 & b = $± \sqrt{2}$

_____________________________________

Answer 2

Answer:

zero in not equal to a

Explanation:

p(x) putting the value of x is (0)

p(x) = x3-3x2+x+1

= 0*3-3*02+0+1

= 1