Question

Find one value of x that is a solution to the equation (x^2+3)^2=4x^2+12

Answer 1

Answer:

${( {x}^{2} + 3) }^{2} = {4x}^{2} + 12$

$\bf \purple{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }$

${( {x}^{2}) }^{2} + 2({x)}^{2} (3) + {(3)}^{2} = {4x}^{2} + 12$

${x}^{4} + {6x}^{2} + 9 = {4x}^{2} + 12$

${x}^{4} = {4x}^{2} - {6x}^{2} + 12 - 9$

${x}^{4} = { - 2x}^{2} + 3$

$( {x}^{2} \times {x}^{2}) = { - 2x}^{2} + 3$

$2 \times {x}^{2} = { - 2x}^{2} + 3$

$= \frac{ { - 2x}^{2} + 3 }{ {2x}^{2} }$

$= 3$