Math, asked by amananil606pcql7e, 1 year ago

if alpha and beta are the zeros of 2x2+5x+k satisfying the relation alpha square+beta square+alpha.beta=21/4,then find the value of k

Answers

Answered by BEJOICE
681

from \:  \: the \:  \: polynomial \\  \alpha  +  \beta  =  \frac{ - 5}{2}  \:  \:  \:  \:  \:  \:  \:  \:  \alpha  \beta  =  \frac{k}{2}  \\given \:  \:  \:  \:   { \alpha }^{2}  +  { \beta }^{2}  +  \alpha  \beta  =  \frac{21}{4}  \\  {( \alpha  +  \beta )}^{2}  -  \alpha  \beta  =  \frac{21}{4}  \\  {( \frac{ - 5}{2} )}^{2}  -  \frac{k}{2}  =  \frac{21}{4}  \\  \frac{k}{2}  =  \frac{25}{4}  -  \frac{21}{4}  = 1 \\ k = 2
Answered by SerenaBochenek
331

Answer:

The value of k is 2

Step-by-step explanation:

Given that if alpha and beta are the zeros of 2x^2+5x+k satisfying the relation

\alpha^2+\beta^2+\alpha.\beta=\frac{21}{4}

we have to find the value of k

Comparing  2x^2+5x+k  with standard quadratic polynomial of form ax^2+bx+c , we get

a=2, b=5 and c=k

\text{Sum of roots=}\alpha+\beta=\frac{-b}{a}=\frac{-5}{2}

\text{Product of roots=}\alpha.\beta=\frac{c}{a}=\frac{k}{2}

\alpha^2+\beta^2+\alpha.\beta=\frac{21}{4}

(\alpha+\beta)^2-\alpha.\beta=\frac{21}{4}

(\frac{-5}{2})^2-\frac{k}{2}=\frac{21}{4}

\frac{25}{4}-\frac{k}{2}=\frac{21}{4}

\frac{25}{4}-\frac{21}{4}=\frac{k}{2}

k=2

The value of k is 2

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