Question

If zeroes of polynomial x^2-3x^2+X+4 are a-b,a and a+b

Answer 1

Answer:

a = 1

b = ±√2

Step-by-step explanation:

p(x)=x

3

−3x

2

+x+1

A standard cubic equation is of the form ax

3

+bx

2

+cx+d

On comparing the two equations, we get

a=1

b=−3

c=1

d=1

Given zeroes are:

α=a−b

β=a

γ=a+b

We know that in a cubic polynomial,

sum of zeroes = −

a

b

⟹α+β+γ=−

1

(−3)

⟹a−b+a+a+b=3

⟹3a=3

\begin{gathered}\implies a = 3 \div 3 = 1\\\\\implies \boxed{\underline{\boxed{\bold{a = 1}}}}\\\\\\\text{We also know that in a cubic polynomial,}\\\\\text{Product of zeroes = $-\dfrac{c}{a}$}\\\\\\\implies \alpha \times \beta \times \gamma = -\dfrac{c}{a}\\\\\\\implies (a-b)(a)(a+b) = -\dfrac{1}{1}\\\\\\\text{Put a = 1}\\\\(1-b)(1)(1+b) = -1\\\\\implies 1^2 - b^2 = -1\\\\\implies 1 - b^2 = -1\\\\\implies b^2 = 1 + 1\\\\\implies b^2 = 2\\\end{gathered}

⟹a=3÷3=1

⟹

a=1

We also know that in a cubic polynomial,

Product of zeroes = −

a

c

⟹α×β×γ=−

a

c

⟹(a−b)(a)(a+b)=−

1

1

Put a = 1

(1−b)(1)(1+b)=−1

⟹1

2

−b

2

=−1

⟹1−b

2

=−1

⟹b

2

=1+1

⟹b

2

=2

\implies \boxed{\underline{\boxed{\bold{b = \pm \sqrt{2}}}}}⟹