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

Let's solve each of the given equations:<br /><br />1. $-3x^{2}+2=0$<br /><br />To solve this equation, we can rearrange it to isolate $x^2$:<br /><br />$-3x^{2} = -2$<br /><br />Dividing both sides by -3:<br /><br />$x^{2} = \frac{2}{3}$<br /><br />Taking the square root of both sides:<br /><br />$x = \pm \sqrt{\frac{2}{3}}$<br /><br />Simplifying the square root:<br /><br />$x = \pm \frac{\sqrt{6}}{3}$<br /><br />Therefore, the solutions to the equation $-3x^{2}+2=0$ are $x = \frac{\sqrt{6}}{3}$ and $x = -\frac{\sqrt{6}}{3}$.<br /><br />2. $2^{\circ }x^{4}+x^{2}-2=0$<br /><br />This equation is not in a standard polynomial form. Let's rewrite it as:<br /><br />$x^{4} + x^{2} - 2 = 0$<br /><br />To solve this equation, we can use substitution. Let $y = x^2$. Then the equation becomes:<br /><br />$y^2 + y - 2 = 0$<br /><br />This is a quadratic equation in $y$. We can solve it using the quadratic formula:<br /><br />$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a = 1$, $b = 1$, and $c = -2$. Plugging in these values:<br /><br />$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(-2)}}{2(1)}$<br /><br />Simplifying:<br /><br />$y = \frac{-1 \pm \sqrt{1 + 8}}{2}$<br /><br />$y = \frac{-1 \pm \sqrt{9}}{2}$<br /><br />$y = \frac{-1 \pm 3}{2}$<br /><br />This gives us two solutions for $y$:<br /><br />$y = 1$ and $y = -2$<br /><br />Since $y = x^2$, we have:<br /><br />$x^2 = 1$ or $x^2 = -2$<br /><br />The first equation gives us $x = \pm 1$, and the second equation has no real solutions since the square of a real number cannot be negative.<br /><br />Therefore, the solutions to the equation $2^{\circ }x^{4}+x^{2}-2=0$ are $x = 1$ and $x = -1$.<br /><br />3. $3^{\circ }4x^{4}-21x^{2}+27=0$<br /><br />This equation is not in a standard polynomial form. Let's rewrite it as:<br /><br />$4x^{4} - 21x^{2} + 27 = 0$<br /><br />To solve this equation, we can use substitution. Let $y = x^2$. Then the equation becomes:<br /><br />$4y^2 - 21y + 27 = 0$<br /><br />This is a quadratic equation in $y$. We can solve it using the quadratic formula:<br /><br />$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br /><br />where $a = 4$, $b = -21$, and $c = 27$. Plugging in these values:<br /><br />$y = \frac{21 \pm \sqrt{(-21)^2 - 4(4)(27)}}{2(4)}$<br /><br />Simplifying:<br /><br />$y = \frac{21 \pm \sqrt{441 - 432}}{8}$<br /><br />$y = \frac{21 \pm \sqrt{9}}{8}$<br /><br />$y = \frac{21 \pm 3}{8}$<br /><br />This gives us two solutions for $y$:<br /><br />$y = \frac{24}{8} = 3$ and $y = \frac{18}{8} = \frac{9}{4}$<br /><br />Since $y = x^2$, we have:<br /><br />$x^2 = 3$ or $x^2 = \frac{9}{4}$<br /><br />Taking the square root of both sides:<br /><br />$x = \pm \sqrt{3}$ or $x = \pm \frac{3}{2}$<br /><br />Therefore, the solutions to the equation $3^{\circ }4x^{4}-21x^{2}+27=0$ are $x = \pm \sqrt{3}$ and $x = \pm \frac{3}{2}$.