Question

If alpha and Beta are the zeros of the quadratic polynomial f(x)= x2 - 5x + k such that alpha - Beta= 1, find the value of k.​

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{ \alpha \: and \: \beta \: are \: zeroes \: of \: f(x)) = {x}^{2} - 5x + k } \\ &\sf{such \: that}\\ &\sf{ \alpha - \beta = 1 } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\: find - \begin{cases} &\sf{value \: of \: k}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline{\bold{Solution :- }}$

Now,

Let us consider a quadratic polynomial f(x) = ax² + bx + c having zeroes α and β, then

$\boxed{{\tt Sum\ of\ the\ zeroes=\frac{-b}{a}}}$

OR

$\boxed{{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

And

$\boxed{{\tt Product\ of\ the\ zeroes=\frac{c}{a}}}$

OR

$\boxed{{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

CALCULATION :-

• The given quadratic polynomial is p(x) = x² - 5x + k

On comparing with

• quadratic polynomial ax² + bx + c,

we get

• a = 1

• b = - 5

• c = k

$\rm\rightarrow \alpha + \beta = \dfrac{-b}{a} = - \dfrac{( - 5)}{1} = 5$

$\rm \: \rightarrow \alpha \beta = \dfrac{c}{a} \: = \dfrac{k}{1} = k$

Now,

$\large \underline{\tt \:{ According \: to \: statement }}$

$\rm :\longmapsto\: \alpha - \beta = 1$

$\rm :\longmapsto\: {( \alpha - \beta )}^{2} = 1$

$\rm :\longmapsto\: {( \alpha + \beta) }^{2} - 4 \alpha \beta = 1$

$\: \: \: \: \: \: \: \: \: \: \{\because \: {(x - y)}^{2} = {(x + y)}^{2} - 4xy \}$

$\rm :\longmapsto\: {5}^{2} - 4k = 1$

$\rm :\longmapsto\:25 - 4k = 1$

$\rm :\longmapsto\:4k = 24$

$\rm :\longmapsto\:k = 6$

Additional Information :-

$\boxed{ \bf \: { \alpha }^{2} + { \beta }^{2} = {( \alpha + \beta )}^{2} - 2 \alpha \beta }$

$\boxed{ \bf { \alpha }^{3} + { \beta }^{3} = {( \alpha + \beta )}^{3} - 3 \alpha \beta ( \alpha + \beta ) }$

$\boxed{ \bf \: {( \alpha - \beta )}^{2} = {( \alpha + \beta )}^{2} - 4 \alpha \beta}$