Question

3/ If the zeroes of the polynomial x - 3x2 + x + 1 are a - b, a, a +b, find a and b.
can any body solbe this so i can check wheather my ans i correct pls
​

Answer 1

Given equation is x³ - 3x² + x + 1

Let alpha = a-b , beta = a, gama = a+b

sum of the roots = -b/a

a - b + a + a - b = -(-3)/1

3a = 3

a = 1

Product of the roots = -d/a

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

(1-b)(1)(1+b) = -1

1 - b² = -1

b² = 2

b = +_√2

Therefore a = 1 , b = +_√2

Answer 2

Answer:

⋆ Given Polynomial : x³ – 3x² + x + 1

Here : a = 1,⠀b = – 3,⠀c = 1,⠀d = 1

• Zeroes are (a – b), a & (a + b)

$\underline{\bigstar\:\textsf{Relation b/w zeroes and coefficient :}}$

$\qquad\underline{\bf{\dag}\:\:\textbf{Sum of Zeroes :}}\\\dashrightarrow\sf\:\:\alpha+\beta+\gamma = \dfrac{-\:b}{a}\\\\\\\dashrightarrow\sf\:\: (a-b)+a+(a+b) = \dfrac{-(-3)}{1}\\\\\\\dashrightarrow\sf\:\:3a=3\\\\\\\dashrightarrow\sf\:\:a=\dfrac{3}{3}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf a=1}}$

$\rule{130}{1}$

$\qquad\underline{\bf{\dag}\:\:\textbf{Product of Zeroes :}}\\\\\dashrightarrow\sf\:\: \alpha \times \beta \times \gamma= \dfrac{-d}{a}\\\\\\\dashrightarrow\sf\:\:(a-b)\times a \times (a+b) = \dfrac{-1}{1}\\\\\\\dashrightarrow\sf\:\:(1-b) \times 1 \times (1+b)=-1\\\\\\\dashrightarrow\sf\:\:\big\lgroup(1)^2-(b)^2\big\rgroup=-1\\\\\\\dashrightarrow\sf\:\:1-b^2=-1\\\\\\\dashrightarrow\sf\:\:1+1=b^2\\\\\\\dashrightarrow\sf\:\:2=b^2\\\\\\\dashrightarrow\sf\:\:\sqrt{2}=b\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf b=\pm\:\sqrt{2}}}$

⠀

$\therefore\:\underline{\textsf{Value of a and b is \textbf{1 \& \pm\sqrt{\text2} } respectively}}.$