Question

solve each equation.

1. $y^{4}-7y^{3}-18y^{2}=0$
2. $m^{4}-625 = 0$
3. $x^{4}-50x^{2}+49 = 0$

Explanation:

Step1: Factor out common factor for equation 13

Factor out $y^{2}$ from $y^{4}-7y^{3}-18y^{2}=0$. We get $y^{2}(y^{2}-7y - 18)=0$.

Step2: Factor quadratic in equation 13

Factor $y^{2}-7y - 18$ as $(y - 9)(y+2)$. So the equation becomes $y^{2}(y - 9)(y + 2)=0$.
Set each factor equal to zero: $y^{2}=0$ gives $y = 0$; $y-9=0$ gives $y=9$; $y + 2=0$ gives $y=-2$.

Step3: Rewrite equation 15

Rewrite $m^{4}-625 = 0$ as $(m^{2})^{2}-25^{2}=0$. Use the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, so $(m^{2}+25)(m^{2}-25)=0$.

Step4: Factor further in equation 15

Factor $m^{2}-25$ as $(m + 5)(m - 5)$. The equation is $(m^{2}+25)(m + 5)(m - 5)=0$.
Set each factor equal to zero: $m^{2}+25=0$ gives $m=\pm5i$; $m + 5=0$ gives $m=-5$; $m - 5=0$ gives $m = 5$.

Step5: Use substitution for equation 17

Let $u=x^{2}$, then the equation $x^{4}-50x^{2}+49=0$ becomes $u^{2}-50u + 49=0$.

Step6: Factor quadratic in equation 17

Factor $u^{2}-50u + 49$ as $(u - 49)(u - 1)=0$.

Step7: Substitute back and solve in equation 17

Substitute $u=x^{2}$ back. We have $x^{2}-49=0$ gives $x=\pm7$ and $x^{2}-1=0$ gives $x=\pm1$.

Answer:

For $y^{4}-7y^{3}-18y^{2}=0$, $y = 0,y=9,y=-2$.
For $m^{4}-625 = 0$, $m = 5,m=-5,m = 5i,m=-5i$.
For $x^{4}-50x^{2}+49=0$, $x = 1,x=-1,x = 7,x=-7$.