Question

$\alpha \: and \: \beta$
are zeroes of the quadratic polynomial
${x}^{2} - (k + 6)x + 2(2k - 1)$
Find the value of k if,
$\alpha + \beta = \frac{ \alpha \beta }{2}$
​

Answer 1

Answer:

                   K = 7

Step-by-step explanation:

Given:

 

              $x^2-(k+6)x+2(2k-1)$

              Here roots are α and β

              For this quadratic equation sum of roots that is ;

              α + β = $\frac{-b}{a}$

              In given quadratic equation a = 1 , b = -(k + 6)

              Therefore : α + β = $\frac{-(-(k+6))}{1}$

              α + β = k + 6 ---------------> Equation 1

              For this quadratic equation multiplication of roots that is ;

               αβ = $\frac{c}{a}$

               In given quadratic equation a = 1 , c = 2(2k-1)

               Therefore :  αβ = $\frac{2(2k -1)}{1}$

               αβ = 2(2k - 1) ------------> Equation 2

               Given:

                                $\alpha +\beta =\frac{\alpha \beta }{2}$

             Substituting equation 1 and 2

              $k+6=\frac{2(2k-1)}{2}$

             k + 6 = 2k - 1

            6 + 1 = 2k - k

            k = 7

                                                   HOPE U UNDERSTOOD

                                                              TAHNK YOU :)

Answer 2

▪ Given :-

• $\sf\alpha \: and \: \beta \: are \: zeroes \: of \\ \sf the \: quadratic \: polynomial \\ \pmb{{x}^{2} - (k + 6)x + 2(2k - 1)}$

• $\pmb{ \alpha + \beta = \dfrac{ \alpha \beta }{2}}$

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▪ To Find :-

• The Value of k.

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▪ Concept To Mind :-

$\large\pmb{\mathfrak{For  \:\text{A} \: Qudratic \: Eq^n  \: of \: the \:Form}}$

$\bf a {x}^{2}  + bx + c$

$\large\pmb{Sum  \: of  \: Zeros  = - \dfrac{b}{a} } \\  \\ \large \pmb{ Product \:  of \:  Zeros =  \frac{c}{a} }$

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▪ Solution :-

$\large \sf{\mathfrak{\text{H} ere ,}}$

$\pmb{Sum  \: of  \: Zeros  = \alpha + \beta =( k + 6)}\\  \\ \pmb{ Product \:  of \:  Zeros = \alpha \beta = 2(2k - 1)}$

$\large \bigstar \underline{ \pmb{ \mathfrak{ \text{A}ccording\:to\:the\:Given\:Condition}}}$

$\alpha + \beta = \dfrac{ \alpha \beta }{2} \\ \\ :\longmapsto k + 6 = \frac{ \cancel2(2k - 1)}{ \cancel2} \\ \\ :\longmapsto k + 6= 2k - 1 \\ \\ \LARGE \purple{ :\longmapsto  \underline {\boxed{{\bf k = 7} }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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