Question

If α and β are the zeroes of the polynomial 2x^2 + 3x + 1 then form the polynomial whose zeroes are α+1 and β+1.

Answer 1

Question:

If α and β are the zeroes of the polynomial 2x^2 + 3x + 1 then form the polynomial whose zeroes are α+1 and β+1.

To find:

$\bigstar To \: find \: the \: equation.$

Answer:

$The \: equation \: is, \underline{ \: \underline{ \sf \red{\bold{2 {x}^{2} - x = 0}}} \: }$

Given:

★ Quadratic equation,

$2 {x}^{2} + 3x + 1 = 0$

$\alpha + 1 \: and \: \beta + 1 \: are \: roots .$

Step-by-step explanation:

$2 {x}^{2} + 3x + 1 = 0$

It's of the form,

$a {x}^{2} + bx + c = 0$

Where a = 2, b = 3, c = 1

$Sum \: of \: roots = \frac{ - b}{a}$

$\boxed{( \alpha + \beta ) = \frac{ - 3}{2} }$

$Product \: of \: roots = \frac{c}{a}$

$\boxed{ \big( \alpha \beta \big) = \frac{1}{2} }$

Now,

Sum of roots,

$( \alpha + 1 ) + ( \beta + 1)$

$\implies (\alpha + \beta) + 2$

Now substituting the values,

$\implies \big( \frac{ - 3}{2} \big) + 2$

$\implies \frac{ - 3 + 4}{2}$

$\implies \boxed{Sum \: of \: roots = \frac{ 1}{2} }$

Product of roots,

$( \alpha + 1 ) ( \beta + 1)$

$\implies \big( \alpha \beta \big) + \big( \alpha + \beta \big) + 1$

Now substituting the values,

$\implies \big( \frac{1}{2} \big) + \big( \frac{ - 3}{2} \big) + 1$

$\implies \big( \frac{ - 2}{2} \big) + 1$

$\implies - 1 + 1$

$\implies \boxed{Product \: of \: roots = 0}$

The equation is formed by,

$\boxed{{x}^{2} - (Sum \: of \: roots)x +Product \: of \: roots = 0}$

$\implies {x}^{2} - \big[\big( \alpha + 1 \big) + \big( \beta + 1 \big) \big]x + \big[\big( \alpha + 1 \big) \big( \beta + 1 \big) \big] = 0$

$\implies {x}^{2} - \frac{1}{2} x + 0 = 0$

$\implies {x}^{2} - \frac{1}{2} x = 0$

Multiplying by 2 on both sides,

$\implies 2 \times {x}^{2} - \not{ 2} \times \frac{1}{ \not{2}} x = 0$

$\implies \boxed{2 {x}^{2} - x = 0}$

$\therefore The \: equation \: is,$

$\underline{ \sf \pink{\bold{2 {x}^{2} - x = 0}}}$