Réitigh sqrt{2x+7}-3x+1=x-1

$$\sqrt { 2 x + 7 } - 3 x + 1 = x - 1$$

Réitigh do x.

$$x = \frac{\sqrt{129} + 9}{16} \approx 1.272363543$$

Céimeanna Réitigh

Bain -3x+1 ón dá thaobh den chothromóid.

$$\sqrt{2x+7}=x-1-\left(-3x+1\right)$$

Chun an mhalairt ar -3x+1 a aimsiú, aimsigh an mhalairt ar gach téarma.

$$\sqrt{2x+7}=x-1-\left(-3x\right)-1$$

Tá 3x urchomhairleach le -3x.

$$\sqrt{2x+7}=x-1+3x-1$$

Comhcheangail x agus 3x chun 4x a fháil.

$$\sqrt{2x+7}=4x-1-1$$

Dealaigh 1 ó -1 chun -2 a fháil.

$$\sqrt{2x+7}=4x-2$$

Cearnaigh an dá thaobh den chothromóid.

$$\left(\sqrt{2x+7}\right)^{2}=\left(4x-2\right)^{2}$$

Ríomh cumhacht \sqrt{2x+7} de 2 agus faigh 2x+7.

$$2x+7=\left(4x-2\right)^{2}$$

Úsáid an teoirim dhéthéarmach \left(a-b\right)^{2}=a^{2}-2ab+b^{2} chun \left(4x-2\right)^{2} a leathnú.

$$2x+7=16x^{2}-16x+4$$

Bain 16x^{2} ón dá thaobh.

$$2x+7-16x^{2}=-16x+4$$

Cuir 16x leis an dá thaobh.

$$2x+7-16x^{2}+16x=4$$

Comhcheangail 2x agus 16x chun 18x a fháil.

$$18x+7-16x^{2}=4$$

Bain 4 ón dá thaobh.

$$18x+7-16x^{2}-4=0$$

Dealaigh 4 ó 7 chun 3 a fháil.

$$18x+3-16x^{2}=0$$

Is féidir gach cothromóid san fhoirm ax^{2}+bx+c=0 a réiteach ag baint úsáid as an bhfoirmle chearnach : \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Tugann an fhoirmle chearnach dhá réiteach, ceann amháin nuair is suimiú é ± agus ceann eile nuair is dealú é.

$$-16x^{2}+18x+3=0$$

Tá an chothromóid seo i bhfoirm chaighdeánach: ax^{2}+bx+c=0. Cuir -16 in ionad a, 18 in ionad b, agus 3 in ionad c san fhoirmle chearnach, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-16\right)\times 3}}{2\left(-16\right)}$$

Cearnóg 18.

$$x=\frac{-18±\sqrt{324-4\left(-16\right)\times 3}}{2\left(-16\right)}$$

Méadaigh -4 faoi -16.

$$x=\frac{-18±\sqrt{324+64\times 3}}{2\left(-16\right)}$$

Méadaigh 64 faoi 3.

$$x=\frac{-18±\sqrt{324+192}}{2\left(-16\right)}$$

Suimigh 324 le 192?

$$x=\frac{-18±\sqrt{516}}{2\left(-16\right)}$$

Tóg fréamh chearnach 516.

$$x=\frac{-18±2\sqrt{129}}{2\left(-16\right)}$$

Méadaigh 2 faoi -16.

$$x=\frac{-18±2\sqrt{129}}{-32}$$

Réitigh an chothromóid x=\frac{-18±2\sqrt{129}}{-32} nuair is ionann ± agus plus. Suimigh -18 le 2\sqrt{129}?

$$x=\frac{2\sqrt{129}-18}{-32}$$

Roinn -18+2\sqrt{129} faoi -32.

$$x=\frac{9-\sqrt{129}}{16}$$

Réitigh an chothromóid x=\frac{-18±2\sqrt{129}}{-32} nuair is ionann ± agus míneas. Dealaigh 2\sqrt{129} ó -18.

$$x=\frac{-2\sqrt{129}-18}{-32}$$

Roinn -18-2\sqrt{129} faoi -32.

$$x=\frac{\sqrt{129}+9}{16}$$

Tá an chothromóid réitithe anois.

$$x=\frac{9-\sqrt{129}}{16} x=\frac{\sqrt{129}+9}{16}$$

Cuir \frac{9-\sqrt{129}}{16} in ionad x sa chothromóid \sqrt{2x+7}-3x+1=x-1.

$$\sqrt{2\times \frac{9-\sqrt{129}}{16}+7}-3\times \frac{9-\sqrt{129}}{16}+1=\frac{9-\sqrt{129}}{16}-1$$

Simpligh. Níl an chothromóid comhlíonann luach x=\frac{9-\sqrt{129}}{16} toisc nach bhfuil ag an taobh clé agus an taobh deas comharthaí os a chomhair.

$$-\frac{15}{16}+\frac{7}{16}\times 129^{\frac{1}{2}}=-\frac{7}{16}-\frac{1}{16}\times 129^{\frac{1}{2}}$$

Cuir \frac{\sqrt{129}+9}{16} in ionad x sa chothromóid \sqrt{2x+7}-3x+1=x-1.

$$\sqrt{2\times \frac{\sqrt{129}+9}{16}+7}-3\times \frac{\sqrt{129}+9}{16}+1=\frac{\sqrt{129}+9}{16}-1$$

Simpligh. An luach x=\frac{\sqrt{129}+9}{16} shásaíonn an gcothromóid.

$$-\frac{7}{16}+\frac{1}{16}\times 129^{\frac{1}{2}}=\frac{1}{16}\times 129^{\frac{1}{2}}-\frac{7}{16}$$

Ag an chothromóid \sqrt{2x+7}=4x-2 réiteach uathúil.

$$x=\frac{\sqrt{129}+9}{16}$$