Question

If the zeroes of the polynomial
$x3 - 3 {x}^{2} + x + 1$
are a-b, a, a+b, find a and b?​

Answer 1

Answer:

a = 1

b = ±√2

Step-by-step explanation:

$$\begin{lgathered}p(x) = x^3 - 3x^2 + x + 1\\\\\text{A standard cubic equation is of the form $ax^3 + bx^2 + cx + d$}\\\\\text{On comparing the two equations, we get}\\\\a = 1\\\\b = -3\\\\c= 1\\\\\\d=1\\\\\text{Given zeroes are:}\\\\\alpha = a-b\\\\\beta = a\\\\\gamma = a + b\\\\\text{We know that in a cubic polynomial,}\\\\\text{sum of zeroes = $-\dfrac{b}{a}$}\\\\\\\implies \alpha + \beta + \gamma = -\dfrac{(-3)}{1}\\\\\\\implies a - b + a + a + b = 3\\\\\implies 3a = 3\\\end{lgathered}$$

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

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