If the zeroes of the polynomial x
^3 – 3x^2+ x + 1 are a – b, a, a + b, then find the value
of a and b.
Answers
Answer:
Answer:
a = 1
b = ±√2
Step-by-step explanation:
\begin{gathered}p(x) = x^3 - 3x^2 + x + 1\\\\\text{A standard cubic equation is of the form $ax^3 + bx^2 + cx + d$}\\\\\text{On comparing the two equations, we get}\\\\a = 1\\\\b = -3\\\\c= 1\\\\\\d=1\\\\\text{Given zeroes are:}\\\\\alpha = a-b\\\\\beta = a\\\\\gamma = a + b\\\\\text{We know that in a cubic polynomial,}\\\\\text{sum of zeroes = $-\dfrac{b}{a}$}\\\\\\\implies \alpha + \beta + \gamma = -\dfrac{(-3)}{1}\\\\\\\implies a - b + a + a + b = 3\\\\\implies 3a = 3\\\end{gathered}
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}}}}}⟹
b=±
2