Question

X^4-3x^2+2=0 then what is the value of x

Answer 1

Answer:

x=±√1,±√2

Step-by-step explanation:

$\sf\{x}^{4} - {3x}^{2} + 2 = 0 \\ \\ \bf\ \implies: { {x}^{2} }^{2} - {3x}^{2} + 2 = 0 \\ \\ \sf\underline{let \: {x}^{2} = y } \\ \\ \sf\ \thereforen {y}^{2} - 3y + 2 = 0 \\ \\\sf\ \implies: {y}^{2} - 2y - y + 2 = 0 \\ \\ \sf\ \implies:y(y - 2) - 1(y - 2) = 0 \\ \\ \bf\ \implies:(y - 2)(y - 1) = 0 \\ \\ \\ \bf\ y - 2 = 0 \\\sf\ \implies:y = 2 \\ \bf\ \implies: {x}^{2} = 2 \\ \bf\ \implies:x = ±\:\sqrt{2} \\ \\ \bf\ y - 1 = 0\\ \bf\ \implies:y = 1\\ \bf\ \implies: {x}^{2} = 1 \\ \bf\ \implies:x = ±\:\sqrt{1}$

Answer 2

Answer:

$\tt\ x = ± \sqrt{1} \: and \: ± \sqrt{2}$

Step-by-step explanation:

$\bf\ {x}^{4} - {3x}^{2} + 2 = 0 \\ \\ \sf\ \implies: { {x}^{2} }^{2} - {3x}^{2} + 2 = 0 \\ \\ \tt\underline{let \: {x}^{2} = y } \\ \\ \sf\ \thereforen {y}^{2} - 3y + 2 = 0 \\ \\\sf\ \implies: {y}^{2} - 2y - y + 2 = 0 \\ \\ \sf\ \implies:y(y - 2) - 1(y - 2) = 0 \\ \\ \sf\ \implies:(y - 2)(y - 1) = 0 \\ \\ \\ \sf\ y - 2 = 0 \\\sf\ \implies:y = 2 \\ \sf\ \implies: {x}^{2} = 2 \\ \sf\ { \implies:x = ±\:\sqrt{2} } \\ \\ \bf\ y - 1 = 0\\ \bf\ \implies:y = 1\\ \bf\ \implies: {x}^{2} = 1 \\ \bf\ { \implies: \:x = ±\:\sqrt{1}}$