Question 1207633
<H3>QUESTION 1</H3>

Let {{{ u=x^2 }}}, so {{{ x^4 + x^2 * sqrt(2) - 2 = 0 }}} becomes {{{ u^2 + sqrt(2) u - 2 = 0 }}}.


Solve for u using the quadratic formula, where {{{ a = 1 }}}, {{{ b = sqrt(2) }}} and {{{ c = -2 }}}:


{{{ u = (-sqrt(2) +- sqrt( (sqrt(2))^2-4*1*(-2) ))/(2*1) }}}


{{{ u = (-sqrt(2) +- sqrt(10))/2 }}}


Substitute back in {{{ u=x^2 }}}:


{{{ x^2 = (-sqrt(2) +- sqrt(10))/2 }}}


As we cannot take the square root of a negative number, the only valid real solution is:


{{{ x^2 = (-sqrt(2) + sqrt(10))/2 }}}


Solving for x gives:


{{{ x = sqrt((-sqrt(2) + sqrt(10))/2) = highlight(0.93) }}}


{{{ x = -sqrt((-sqrt(2) + sqrt(10))/2) = highlight(-0.93) }}}



<H3>QUESTION 2</H3>

Rearrange {{{ pi(1 + r)^2 = 2 + pi(1 + r) }}} so that it is in the form {{{ ax^2 + bx + c = 0 }}}:


{{{ pi(1 + r)^2 - pi(1 + r) - 2 = 0 }}}


Expand:


{{{ pi*(1 + 2r + r^2) - pi - pi * r - 2 = 0 }}}

{{{ pi + 2 * pi * r + pi * r^2 - pi - pi * r - 2 = 0 }}}

{{{ pi * r^2 + pi * r - 2 = 0}}}


Solve for r using the quadratic formula, where {{{ a = pi }}}, {{{ b = pi }}} and {{{ c = -2 }}}:


{{{ r = (-pi +- sqrt( (pi)^2-4*pi*(-2) ))/(2*pi) }}}


{{{ r = (-pi +- sqrt( (pi)^2+8*pi))/(2*pi) }}}


Solve for r:


{{{ r = (-pi + sqrt( (pi)^2+8*pi))/(2*pi) = highlight(0.44) }}}


{{{ r = (-pi - sqrt( (pi)^2+8*pi))/(2*pi) = highlight(-1.44) }}}