Réitigh (2y+3)^2+y^2=4

$$( 2 y + 3 ) ^ { 2 } + y ^ { 2 } = 4$$

Réitigh do y.

$$y=\frac{\sqrt{11}-6}{5}\approx -0.536675042$$

$$y=\frac{-\sqrt{11}-6}{5}\approx -1.863324958$$

Céimeanna ag Baint Úsáid as an bhFoirmle Chearnach

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(2y+3\right)^{2} a leathnú.

$$4y^{2}+12y+9+y^{2}=4$$

Comhcheangail 4y^{2} agus y^{2} chun 5y^{2} a fháil.

$$5y^{2}+12y+9=4$$

Bain 4 ón dá thaobh.

$$5y^{2}+12y+9-4=0$$

Dealaigh 4 ó 9 chun 5 a fháil.

$$5y^{2}+12y+5=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir 5 in ionad a, 12 in ionad b, agus 5 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-12±\sqrt{12^{2}-4\times 5\times 5}}{2\times 5}$$

Cearnóg 12.

$$y=\frac{-12±\sqrt{144-4\times 5\times 5}}{2\times 5}$$

Méadaigh -4 faoi 5.

$$y=\frac{-12±\sqrt{144-20\times 5}}{2\times 5}$$

Méadaigh -20 faoi 5.

$$y=\frac{-12±\sqrt{144-100}}{2\times 5}$$

Suimigh 144 le -100?

$$y=\frac{-12±\sqrt{44}}{2\times 5}$$

Tóg fréamh chearnach 44.

$$y=\frac{-12±2\sqrt{11}}{2\times 5}$$

Méadaigh 2 faoi 5.

$$y=\frac{-12±2\sqrt{11}}{10}$$

Réitigh an chothromóid y=\frac{-12±2\sqrt{11}}{10} nuair is ionann ± agus plus. Suimigh -12 le 2\sqrt{11}?

$$y=\frac{2\sqrt{11}-12}{10}$$

Roinn -12+2\sqrt{11} faoi 10.

$$y=\frac{\sqrt{11}-6}{5}$$

Réitigh an chothromóid y=\frac{-12±2\sqrt{11}}{10} nuair is ionann ± agus míneas. Dealaigh 2\sqrt{11} ó -12.

$$y=\frac{-2\sqrt{11}-12}{10}$$

Roinn -12-2\sqrt{11} faoi 10.

$$y=\frac{-\sqrt{11}-6}{5}$$

Tá an chothromóid réitithe anois.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$

Céimeanna chun an Chearnóg a Chomhlánú

Úsáid an teoirim dhéthéarmach \left(a+b\right)^{2}=a^{2}+2ab+b^{2} chun \left(2y+3\right)^{2} a leathnú.

$$4y^{2}+12y+9+y^{2}=4$$

Comhcheangail 4y^{2} agus y^{2} chun 5y^{2} a fháil.

$$5y^{2}+12y+9=4$$

Bain 9 ón dá thaobh.

$$5y^{2}+12y=4-9$$

Dealaigh 9 ó 4 chun -5 a fháil.

$$5y^{2}+12y=-5$$

Roinn an dá thaobh faoi 5.

$$\frac{5y^{2}+12y}{5}=-\frac{5}{5}$$

Má roinntear é faoi 5 cuirtear an iolrúchán faoi 5 ar ceal.

$$y^{2}+\frac{12}{5}y=-\frac{5}{5}$$

Roinn -5 faoi 5.

$$y^{2}+\frac{12}{5}y=-1$$

Roinn \frac{12}{5}, comhéifeacht an téarma x, faoi 2 chun \frac{6}{5} a fháil. Ansin suimigh uimhir chearnach \frac{6}{5} leis an dá thaobh den chothromóid. Déanann an chéim seo slánchearnóg de thaobh clé na cothromóide.

$$y^{2}+\frac{12}{5}y+\left(\frac{6}{5}\right)^{2}=-1+\left(\frac{6}{5}\right)^{2}$$

Cearnaigh \frac{6}{5} trí uimhreoir agus ainmneoir an chodáin a chearnú.

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=-1+\frac{36}{25}$$

Suimigh -1 le \frac{36}{25}?

$$y^{2}+\frac{12}{5}y+\frac{36}{25}=\frac{11}{25}$$

Fachtóirigh y^{2}+\frac{12}{5}y+\frac{36}{25}. Go ginearálta, nuair x^{2}+bx+c cearnóg fhoirfe é, is féidir é a fhachtóiriú i gcónaí mar \left(x+\frac{b}{2}\right)^{2}.

$$\left(y+\frac{6}{5}\right)^{2}=\frac{11}{25}$$

Tóg fréamh chearnach an dá thaobh den chothromóid.

$$\sqrt{\left(y+\frac{6}{5}\right)^{2}}=\sqrt{\frac{11}{25}}$$

Simpligh.

$$y+\frac{6}{5}=\frac{\sqrt{11}}{5} y+\frac{6}{5}=-\frac{\sqrt{11}}{5}$$

Bain \frac{6}{5} ón dá thaobh den chothromóid.

$$y=\frac{\sqrt{11}-6}{5} y=\frac{-\sqrt{11}-6}{5}$$