11 Solve each of the following bisquared equations : 10 x^4-3x^2+2=0 20 x^4+x^2-2=0 3^circ 4x^4-21x^2+27=0

To solve the, we can use substitution to simplify the process. Let's solve each equation step by step:<br /><br />1. $x^{4}-3x^{2}+2=0$<br /><br />Let $y = x^2$. Then the equation becomes $y^2 - 3y + 2 = 0$. This is a quadratic equation in $y$. We can solve it by factoring:<br /><br />$(y - 1)(y - 2) = 0$<br /><br />So, $y = 1$ or $y = 2$. Now, substitute back $y = x^2$:<br /><br />$x^2 = 1$ or $x^2 = 2$<br /><br />Therefore, $x = \pm 1$ or $x = \pm \sqrt{2}$.<br /><br />2. $x^{4}+x^{2}-2=0$<br /><br />Let $y = x^2$. Then the equation becomes $y^2 + y - 2 = 0$. This is a quadratic equation in $y$. We can solve it by factoring:<br /><br />$(y + 2)(y - 1) = 0$<br /><br />So, $y = -2$ or $y = 1$. Now, substitute back $y = x^2$:<br /><br />$x^2 = -2$ or $x^2 = 1$<br /><br />Therefore, $x = \pm i\sqrt{2}$ or $x = \pm 1$.<br /><br />3. $3x^{4}-21x^{2}+27=0$<br /><br />Let $y = x^2$. Then the equation becomes $3y^2 - 21y + 27 = 0$. We can simplify this equation by dividing all terms by 3:<br /><br />$y^2 - 7y + 9 = 0$<br /><br />Now, we can solve this quadratic equation using the quadratic formula:<br /><br />$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a = 1$, $b = -7$, and $c = 9$.<br /><br />$y = \frac{7 \pm \sqrt{49 - 36}}{2}$<br /><br />$y = \frac{7 \pm \sqrt{13}}{2}$<br /><br />So, $y = \frac{7 + \sqrt{13}}{2}$ or $y = \frac{7 - \sqrt{13}}{2}$.<br /><br />Now, substitute back $y = x^2$:<br /><br />$x^2 = \frac{7 + \sqrt{13}}{2}$ or $x^2 = \frac{7 - \sqrt{13}}{2}$<br /><br />Therefore, $x = \pm \sqrt{\frac{7 + \sqrt{13}}{2}}$ or $x = \pm \sqrt{\frac{7 - \sqrt{13}}{2}}$.