Math, asked by rasel7345, 10 months ago

If alpha and beta are zeroes of 2x2+5x+k such that alpha2+beta2 + alpha beta = 21/4

Answers

Answered by Sudhir1188
10

Question should be:

  • If alpha and beta are zeroes of 2x²+5x+k such that (alpha)²+(beta)² + (alpha)(beta) = 21/4

ANSWER:

  • Value of k is 2

GIVEN:

  • P(x) = 2x²+5x+k
  • α²+β²+αβ = 21/4

TO FIND:

  • Value of 'k'.

SOLUTION:

P(x) = 2x²+5x+k

Here:

Coefficient of x² = 2

Coefficient of x = 5

Constant term = k

Formula:

=> Sum of zeroes (α+β) = -(Coefficient of x)/Coefficient of x²

=> Product of zeroes (αβ) = Constant term/ Coefficient of x²

=> Sum of zeros = -5/2

=> Product of zeros = k/2

Now:

 \implies \:  \alpha {}^{2}  +  \beta {}^{2}  +  \alpha \beta \:  =  \dfrac{21}{4}  \\  \\  \implies \: ( \alpha \:  +  \beta) {}^{2}  -  \alpha \beta \:  =  \dfrac{21}{4}  \\  \\  \:  \:  \: putting \: the \: values \: we \: get. \\  \implies \: ( \dfrac{ - 5}{2} ) {}^{2}  -  \dfrac{k}{2}  =  \dfrac{21}{4}  \\  \\  \implies \:  \dfrac{25}{4}  -  \dfrac{21}{4}  =  \dfrac{k}{2}  \\  \\  \implies \:  \dfrac{4}{4}  =  \dfrac{k}{2}  \\  \\  \implies \: k \:  = 2

Value of k is 2.

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