if alpha and beeta are zeroes of quadratic polynomial 2x²+5+k , find the value of k such that (alpha+beeta)²-alpha beetta =24
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Answers
Answer:
2x² + 5x + k = 0 is the given quadratic
equation and it is given that (a+B)² - a.ß = 24 ....(i)
As we know that,
a+ß = -b/a = -5/2 ....(ii) and a.ß = c/a = k/2 ....(iii)
Putting values of (ii) & (iii) in (i), we get
(-5/2)²-(k/2) = 24
=> 25/4 - k/2 = 24 => (25-2k)/4 = 24
=> 25 - 2k = 96
=> 2k = 25-96
=> 2k = -71
=> k = -71/2
hope it helps ( ¬¬)
happy to help :)
Answer:
2x² + 5x + k = 0 is the given quadratic
equation and it is given that (a+B)² - a.ß = 24 ....(i)
As we know that,
a+ß = -b/a = -5/2 ....(ii) and a.ß = c/a = k/2 ....(iii)
Putting values of (ii) & (iii) in (i), we get
(-5/2)²-(k/2) = 24
=> 25/4 - k/2 = 24 => (25-2k)/4 = 24
=> 25 - 2k = 96
=> 2k = 25-96
=> 2k = -71
=> k = -71/2
happy to help :)
Step-by-step explanation:
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