Question

Determinati x∈R pentru care4(2x-5)²-(3x+1)²=0.Va rog frumos :*

Answer 1

4(2x-5)²-(3x+1)²=0⇒ (8x-20)²-(3x+1)²=0⇒64x²-320x+400-(9x²+6x+1)=0⇒64x²-320x+400-9x²-6x-1=0⇒55x²+314x+399=0

Δ=314²-4*55*399=98596-87780=10816

x(1)=(-314+√10816)/110⇒X(1)=(-314+104)/110⇒x(1)=-210/110⇒x(1)=-21/11

x(2)=(-314-√10816)/110⇒x(2)=(-314-104)/110⇒x(2)=-418/110⇒x(2)=-209/55

Answer 2

4(2x-5)²-(3x+1)²=0
(2x-5)=(2x)²+2*2x*5+5²
(3x+1)²=(3x)²+2*3x*1+1²
AM CALC. DUPA FORMULA (a+b)²=a²+2ab+b²  se poate si cu -.⇒a²-2ab+b².
⇒4x²+20x+25-(minusul schimba semnul) 9x²-6x-1=0
⇒5x²+14x+24=0
apoi faci un artificiu de calcul.