Question

Determinați XeR astfel încât
$4(x + 3) ^{2} - 7 = 29$
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Answer 1

Răspuns:

x₁ = 0

x₂ = -6

Explicație pas cu pas:

4(x+3)² = 36

(x+3)² = 9

x+3 = ± 3

a) x+3 = 3  ⇒ x = 0

b) x+3 = -3 ⇒ x = -6

Answer 2

$\bf \blue{(a + b) {}^{2} = {a}^{2} + 2ab + {b}^{2} }$

$4(x + 3) {}^{2} - 7 = 29 \\ 4(x {}^{2} + 2 \times x \times 3 + 3 {}^{2} ) - 7 - 29 = 0 \\ 4( {x}^{2} + 6x + 9) - 36 = 0 \\ 4x {}^{2} + 4 \times 6x + 4 \times 9 - 36 = 0 \\ 4x {}^{2} + 24x + 36 - 36 = 0 \\ 4x {}^{2} + 24x = 0 \\ 4(x {}^{2} + 6x) = 0 | \div 4\\ x {}^{2} + 6x = 0$

{a=1

{b=6

{c=0

delta=b²- 4ac=6²-4•1•0=36-0=36

x1= -b - ✓delta /2a = -6 - ✓36 /2•1 = - 6 -6/2

x1= -12/2 = - 6

x2= -b + ✓delta /2a = -6 + ✓36 /2•1

x2= -6+6/2 = 0/2 =0

x1=-6

x2=0