Question

If ∝, β are the roots of the equation $x^2-ax+b=0$, then ${\alpha }^4$+${\alpha }^3$β+${\alpha }^2$${\beta }^2$+α${\beta }^3$+${\beta }^4$

• A. None of these
• B. $a^3+3a^{2}b$ + $b^2$
• C. $a^4+3a^{2}b$ + $b^2$
• D. $a^4-3a^{2}b$ + $b^2$

Answer 1

The correct option is D

$a^4-3a^{2}b$ + $b^2$

α+β $=a$, αβ $=b$

therefore ${\alpha }^4$+${\alpha }^3$β+${\alpha }^2$${\beta }^2$+α${\beta }^3$+${\beta }^4$

= (α+β) (${\alpha }^3$+${\beta }^3$) + ${\alpha }^2$${\beta }^2$

=${(\alpha +\beta )}^2$[${(\alpha +\beta )}^2-3$αβ]+ ${\left(\alpha \beta \right)}^2$
$=a^2$($a^2–3b)$ +$b^2$
$=a^4–3a^{2}b$ + $b^2$