Question

hw 1 - algebra for calculus section 0.1: problem 3 (1 point)
solve the following equations. if there is more than one solution, enter the solutions as a list of numbers separated by commas.
r(12r + 11)= - 2
r=
10z² + 21z = 10
z=

Explanation:

Step1: Expand first equation

Expand $r(12r + 11)=-2$ to get $12r^{2}+11r + 2=0$.

Step2: Factor the quadratic

Factor $12r^{2}+11r + 2$ as $(3r + 2)(4r+1)=0$.

Step3: Solve for r

Set each factor equal to zero:
For $3r + 2 = 0$, $r=-\frac{2}{3}$; for $4r + 1=0$, $r=-\frac{1}{4}$.

Step4: Rearrange second equation

Rearrange $10z^{2}+21z = 10$ to $10z^{2}+21z - 10=0$.

Step5: Factor the quadratic

Factor $10z^{2}+21z - 10$ as $(5z - 2)(2z+5)=0$.

Step6: Solve for z

Set each factor equal to zero:
For $5z - 2 = 0$, $z=\frac{2}{5}$; for $2z + 5=0$, $z=-\frac{5}{2}$.

Answer:

$r=-\frac{2}{3},-\frac{1}{4}$; $z=\frac{2}{5},-\frac{5}{2}$