Issolvi (2x+1)^2-4(x+1)^2leq-2

$$( 2 x + 1 ) ^ { 2 } - 4 ( x + 1 ) ^ { 2 } \leq - 2$$

Solvi għal x

$$x\geq -\frac{1}{4}$$

Passi tas-Soluzzjonijiet

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(2x+1\right)^{2}.

$$4x^{2}+4x+1-4\left(x+1\right)^{2}\leq -2$$

Uża teorema binomjali \left(a+b\right)^{2}=a^{2}+2ab+b^{2} biex tespandi \left(x+1\right)^{2}.

$$4x^{2}+4x+1-4\left(x^{2}+2x+1\right)\leq -2$$

Uża l-propjetà distributtiva biex timmultiplika -4 b'x^{2}+2x+1.

$$4x^{2}+4x+1-4x^{2}-8x-4\leq -2$$

Ikkombina 4x^{2} u -4x^{2} biex tikseb 0.

$$4x+1-8x-4\leq -2$$

Ikkombina 4x u -8x biex tikseb -4x.

$$-4x+1-4\leq -2$$

Naqqas 4 minn 1 biex tikseb -3.

$$-4x-3\leq -2$$

Żid 3 maż-żewġ naħat.

$$-4x\leq -2+3$$

Żid -2 u 3 biex tikseb 1.

$$-4x\leq 1$$

Iddividi ż-żewġ naħat b'-4. Peress li -4 huwa negattiv, id-direzzjoni tal-inugwaljanza inbidlet.

$$x\geq -\frac{1}{4}$$