Question

If α and β are the zeroes of x²-4x+1, then α² +β² - 9αβ​

Answer 1

${\blue{ \boxed{ \boxed{ \begin{array}{cc} \to \bf \: given \: : \\ \\ \rm \: \alpha \: \: and \: \beta \: \: are \: the \: zeroes \: of \: the \: quadratic \\ \rm \: polynomial \: \: \: \: {x}^{2} - 4x + 1 \\ \\ \\ \red{ \underline{\sf \: we \: have \: to \: find \: the \: value \: of :}} \\ \\ \sf \hookrightarrow \: { \alpha }^{2} + { \beta }^{2} - 9 \alpha \beta \end{array}}}}}$

$\small{{\pink{ \boxed{ \boxed{ \begin{array}{cc} \green{ \underline{ \sf \: we \: know \: that : }} \\ \\ \rm \: if \: \: a {x}^{2} + bx + c \: \: is \: a \: quadratic \\ \rm polynomial \: \rm \: and \: \alpha \: , \: \beta \: are \: the \: zeroes \\ \rm \: then \\ \\ \small{\boxed{\sf \: sum \: of \: zeroes \:, ( \alpha + \beta ) = - \frac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} } = - \frac{b}{a} } } \\ \\ \\ \small{ \boxed{\sf \: product \: of \: zeroes, \: \alpha \beta = \frac{constant \: term}{coefficient \: of \: {x}^{2} } = \frac{c}{a}}} \\ \\ \end{array}}}}}}$

According to the question,

a = 1

b = -4

c = 1

Now,

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: sum \: of \: zeroes , \: \alpha + \beta = - \frac{ - 4}{1} = 4 \\ \\ \sf \: product \: of \: zeroes, \: \alpha \beta = \frac{1}{1} = 1 \end{array}}}}}$

Again,

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \: we \: know \: that \: \\ {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy \end{array}}}}}$

So,

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta \\ \\ = {4}^{2} - 2 \times 1 \\ \\ = 16 - 2 \\ \\ = 14 \end{array}}}}}$

Now evaluating,

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: { \alpha }^{2} + { \beta }^{2} - 9 \alpha \beta \\ \\ = 14 - 9 \times 1 \\ \\ = 14 - 9 \\ \\ = 5 \end{array}}}}}$

So,

${ \alpha }^{2} + { \beta }^{2} - 9 \alpha \beta = 5$