Sagot :
bonjour
je te montre
A (x) = x² + 6 x + 9
d = 36 - 4 ( 1 * 9 ) = 36 - 36 = 0
d = 0 donc une seule solution
x 0 = - 6 / 2 = - 3
A (x) = ( x + 3 ) ( x + 3 )
B ( x) = - x² + 2 x + 15
d = 4 - 4 ( - 1 * 15 )= 4 + 60 = 64
x 1 = ( 2 - 8 ) / - 2 = - 6 / - 2 = 3
x 2 = ( 2 + 8 ) / - 2 = - 5
b (x) = ( x - 3 ) ( x + 5 )
continue
32
a)
A(x) = x² + 6x + 9
A(x) = (x + 3)²
b)
B(x) = -x² + 2x + 15
= - (x² - 2x) + 15
= - (x² - 2x + 1 - 1) + 15
= - [(x - 1)² -1] + 15
= -(x -1)² + 1 + 15
= 16 - (x - 1)²
= 4² - (x -1)²
= [4 - (x - 1)][ 4 + (x - 1)]
= (-x + 5)(x + 3)
= - (x - 5)(x + 3)
45
cas général
f(x) = ax² + bx + c
f(x) = a(x - x1)(x - x2)
1)
f(x) = a(x -2)(x + 8) a non nul
2)
f(x) est de la forme f(x) = a(x + 1/2)(x - 5)
on calcule a en écrivant que f(0) = -10
f(0) = a*(1/2) *(-5) = (-5/2)a
(-5/2)a = -10
a = -10 /(-5/2)
a = 4
f(x) = 4(x + 1/2)(x - 5)
f(x) = 2(2x + 1)(x - 5)
3)
idem