equação A)×2-7×+10=0 B) ×2-8×+7=0 C)-2×2-4×+48=0
Soluções para a tarefa
Equação
A)×2-7×+10=0
x² - 7x + 10 = 0
a = 1
b = - 7
c = 10
Δ = b² - 4ac
Δ = (-7)² - 4(1)(10)
Δ = + 49 - 40
Δ = + 9 ----------------------------------> √Δ = 3( porque √9 = 3)
se
Δ > 0 ( DUAS raizes diferentes)
(baskara)
→→→→(- b + - √Δ)
x = ---------------------
→→→→→→→(2a)
=========================================================
x' = -(-7) - √9/2(1)
x' = + 7 - 3/2
x' = + 4/2
x' = 2
e
x'' = -(-7) + √9/2(1)
x'' = + 7 + 3/2
x'' = + 10/2
x'' = 5
=====================================================
B)
×2-8×+7=0
x² - 8x + 7 = 0
a = 1
b = - 8
c = 7
Δ = b² - 4ac
Δ = (-8)² - 4(1)(7)
Δ = + 64 - 28
Δ = + 36 --------------------------->√Δ = 6 ( porque √36 = 6)
se
Δ > 0 ( DUAS raizes diferentes)
(baskara)
→→→→(- b + - √Δ)
x = ---------------------
→→→→→→→(2a)
=============================================================
x' = -(-8) - √36/2(1)
x' = + 8 - 6/2
x' = + 2/2
x' = 1
e
x'' = -(-8) + √36/2(1)
x'' = + 8 + 6/2
x'' = + 14/2
x'' = 7
=============================================================
C)-2×2-4×+48=0
-2x² - 4x + 48 = 0
a = - 2
b = - 4
c = 48
Δ = b² - 4ac
Δ = (-4)² - 4(-2)(48)
Δ = + 16 + 384
Δ = + 400 -------------------------> √Δ = 20 ( porque √400 = 20)
se
Δ > 0 ( DUAS raizes diferentes)
(baskara)
→→→→(- b + - √Δ)
x = ---------------------
→→→→→→→(2a)
==================================================
x' = -(-4) - √400/2(-2)
x' = + 4 - 20/-4
x' = - 16/-4
x' = + 16/4
x' = + 4
e
x'' = -(-4) + √400/2(-2)
x'' = + 4 + 20/-4
x'' = + 24/-4
x'' = - 24/4
x'' = - 6