Question

If a(alpha) and ß(beta) are the roots of the equation 2x² + Kx + 5 = 0 when k is a constant and a² + ß² = -1 find the value of k​

Answer 1

Answer:

The solution is given above.

Step-by-step explanation:

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Answer 2

Given Polynomial : 2x² + kx + 5 = 0

☆ On Comparing the given polynomial with standard form of Quadratic Equation ( i.e. ax² + bx + c = 0 ) we get —

• a = 2 ,
• b = k &
• c = 5 .

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✇ Now , Finding out the Sum and Product of zeroes of Quadratic Equation . Sum and Product of zeroes are given by —

• ( α + β ) = –b/a
• ( α β ) = c/a

$\\\qquad \dag\: \underline{\pmb{\sf \:Sum\:of\:Roots\:(\:\alpha + \beta \:)\::\:}}$

$\dashrightarrow \sf \Big\{ \alpha + \beta \Big\}\:=\:\dfrac{ - b }{a}\:\\\\\\ \dashrightarrow \sf \Big\{ \alpha + \beta \Big\}\:=\:\dfrac{ - (k)}{2}\:\\\\\\ \dashrightarrow \underline {\boxed { \pmb{\frak{ \alpha + \beta \:=\:\dfrac{ - k }{2}\:}}}}\:\bigstar$

$\qquad \dag\: \underline{\pmb{\sf \:Product \:of\:Roots\:(\:\alpha \beta \:)\::\:}}$

$\dashrightarrow \sf \Big\{ \alpha \beta \Big\}\:=\:\dfrac{ c }{a}\:\\\\\\ \dashrightarrow \sf \Big\{ \alpha \beta \Big\}\:=\:\dfrac{ 5}{2}\:\\\\\\ \dashrightarrow \underline {\boxed { \pmb{\frak{ \alpha \beta \:=\:\dfrac{ 5 }{2}\:}}}}\:\bigstar$

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We're Provided with that α² + β² = – 1 , By using this we get —

$\twoheadrightarrow \:\sf \alpha^2\:+\:\beta^2\:=\:-1\:\\\\\\ \twoheadrightarrow \:\sf (\: \alpha\:+\:\beta\:)^2\:-\:2\alpha\beta\:=\:-1\:$

• ( α + β ) = -k/2
• ( α β ) = 5/2

$\\ \twoheadrightarrow \:\sf (\: \alpha\:+\:\beta\:)^2\:-\:2\alpha\beta\:=\:-1\:\\\\\\ \twoheadrightarrow \:\sf \bigg( \:\dfrac{-k}{2}\:\bigg)^2\:-\:2\bigg( \:\dfrac{5}{2}\:\bigg)\:=\:-1\:\\\\\\ \twoheadrightarrow \:\sf \:\dfrac{k^2}{4}\:\:-\:2\times \:\dfrac{5}{2}\:=\:-1\:\\\\\\ \twoheadrightarrow \:\sf \:\dfrac{k^2}{4}\:\:-\:5\:=\:-1\:\\\\\\ \twoheadrightarrow \:\sf \:k^2\:=\:16\:\\\\\\ \twoheadrightarrow \:\sf \:k\:=\:\pm\:\sqrt{\Big(16\Big)}\:\\\\\\ \twoheadrightarrow \underline {\boxed { \pmb{\frak{ k \:=\:\pm\:4\:}}}}\:\bigstar$

$\therefore \:\underline {\sf Hence, \:Value \:of\:k\:is\:\pmb{\sf \pm\:4\:}\:.}$