Question

solve.

1. $x^{4}-6x^{2}=8$
2. $x^{4}-3x^{2}=54$
3. $m^{6}-16m^{5}+64m^{4}=0$
4. $y^{4}-5y^{3}=-4y^{2}$

Explanation:

Step1: Solve $x^{4}-6x^{2}=8$

Let $u = x^{2}$, then the equation becomes $u^{2}-6u - 8=0$. Using the quadratic - formula $u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $au^{2}+bu + c = 0$. Here, $a = 1$, $b=-6$, $c=-8$.
$u=\frac{6\pm\sqrt{(-6)^{2}-4\times1\times(-8)}}{2\times1}=\frac{6\pm\sqrt{36 + 32}}{2}=\frac{6\pm\sqrt{68}}{2}=\frac{6\pm2\sqrt{17}}{2}=3\pm\sqrt{17}$.
Since $u = x^{2}$, then $x^{2}=3+\sqrt{17}$ or $x^{2}=3 - \sqrt{17}$. But $3-\sqrt{17}<0$, so we discard it for real - valued solutions. For $x^{2}=3+\sqrt{17}$, $x=\pm\sqrt{3 + \sqrt{17}}$.

Step2: Solve $m^{6}-16m^{5}+64m^{4}=0$

Factor out $m^{4}$ first: $m^{4}(m^{2}-16m + 64)=0$.
Then factor the quadratic part: $m^{4}(m - 8)^{2}=0$.
Set each factor equal to zero: $m^{4}=0$ gives $m = 0$ (with multiplicity 4), and $(m - 8)^{2}=0$ gives $m = 8$ (with multiplicity 2).

Step3: Solve $y^{4}-5y^{3}=-4y^{2}$

Move all terms to one side: $y^{4}-5y^{3}+4y^{2}=0$.
Factor out $y^{2}$: $y^{2}(y^{2}-5y + 4)=0$.
Factor the quadratic part: $y^{2}(y - 1)(y - 4)=0$.
Set each factor equal to zero: $y^{2}=0$ gives $y = 0$ (with multiplicity 2), $y - 1=0$ gives $y = 1$, and $y - 4=0$ gives $y = 4$.

Answer:

For $x^{4}-6x^{2}=8$, $x=\pm\sqrt{3+\sqrt{17}}$; for $m^{6}-16m^{5}+64m^{4}=0$, $m = 0$ (multiplicity 4), $m = 8$ (multiplicity 2); for $y^{4}-5y^{3}=-4y^{2}$, $y = 0$ (multiplicity 2), $y = 1$, $y = 4$