Resposta:
x²- (p + 5)x + 36 = 0
a = 1
b = - (p + 5) = - p - 5
c = 36
∆ = 0
b² - 4ac = 0
[-(p+5)}² - 4(1 )( 36) =0
a=1
b=10
c= -119
∆ = b² - 4ac
∆ = 10² - 4.1.(- 119)
∆ = 100 + 476
∆ = 576
[tex]p=\dfrac{-b\pm\sqrt{\Delta} }{2a}\\ \\ \\ p=\dfrac{-10\pm\sqrt{576} }{2(1)}=\dfrac{-10\pm24}2}\\ \\\\ p'=\dfrac{-10-24}{2}=-\dfrac{34}{2}=-17\\ \\ \\ p"=\dfrac{-10+24}{2}=\dfrac{14}{2}=7[/tex]
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Resposta:
x²- (p + 5)x + 36 = 0
a = 1
b = - (p + 5) = - p - 5
c = 36
∆ = 0
b² - 4ac = 0
[-(p+5)}² - 4(1 )( 36) =0
p² + 10p - 119=0
a=1
b=10
c= -119
∆ = b² - 4ac
∆ = 10² - 4.1.(- 119)
∆ = 100 + 476
∆ = 576
[tex]p=\dfrac{-b\pm\sqrt{\Delta} }{2a}\\ \\ \\ p=\dfrac{-10\pm\sqrt{576} }{2(1)}=\dfrac{-10\pm24}2}\\ \\\\ p'=\dfrac{-10-24}{2}=-\dfrac{34}{2}=-17\\ \\ \\ p"=\dfrac{-10+24}{2}=\dfrac{14}{2}=7[/tex]
Logo os valores procurados são p’=-17 e p’’=7