Ufumbuzi - Kutatua kutofautiana za quadratic kwa kutumia formula ya quadratic

Kupata mizizi ya equation ya quadratic, plaga vipingamizi vyake ($$a$$, $$b$$ na $$c$$) kwenye formula ya quadratic:

$$x=\left(-b\pm sqrt\left(b^2-4ac\right)\right)/\left(2a\right)$$

$$x=\left(-1*-9\pm sqrt\left(-9^2-4*1*18\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-9\pm sqrt\left(81-4*1*18\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-9\pm sqrt\left(81-4*18\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-9\pm sqrt\left(81-72\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-9\pm sqrt\left(9\right)\right)/\left(2*1\right)$$

$$x=\left(-1*-9\pm sqrt\left(9\right)\right)/\left(2\right)$$

$$x=\left(9\pm sqrt\left(9\right)\right)/2$$

kupata matokeo:

$$x=\left(9\pm sqrt\left(9\right)\right)/2$$

Rahisisha $$9$$ kwa kugundua vigezo vyake vya asili:

Ugawanyaji wa kihesabu wa $$9$$ ni $$3^2$$

$$x=\left(9\pm 3\right)/2$$

The ± inamaanisha mizizi mbili zinawezekana.

Pekua usawa:
$$x_1=\left(9+3\right)/2$$ na $$x_2=\left(9-3\right)/2$$

$$x_1=\left(9+3\right)/2$$

$$x_1=\left(9+3\right)/2$$

$$x_1=\left(12\right)/2$$

$$x_1=\frac{12}{2}$$

$$x_1=6$$

$$x_2=\left(9-3\right)/2$$

$$x_2=\left(9-3\right)/2$$

$$x_2=\left(6\right)/2$$

$$x_2=\frac{6}{2}$$

$$x_2=3$$

Kupata vipindi vya usawa wa quadratic, tunaanza kwa kupata parabola yake.

Mizizi ya parabola (ambapo inakutana na mhimili wa x) ni: 3, 6.

Kwa sababu idadi ya namba $$a$$ ni chanya ($$a$$=1), hii ni usawa wa "positive" quadratic na parabola inaelekea juu, kama tabasamu!

If ishara ya usawa ni ≤ au ≥ , basi vipindi vinajumuisha mizizi na tunatumia mstari mzito. If ishara ya usawa ni < au > vipindi havijumuishi mizizi na tunatumia mstari wa nukta.

Kwa kuwa $$x^2-9x+18>0$$ ina ishara ya kutofautiana $$>$$, tunatafuta vipindi vya parabola ambavyo ni juu ya mhimili wa x.

Ufumbuzi: $$$$

Notation ya kipindi: $$$$