Solving equations

Printable View

August 2nd 2006, 08:13 PM
pashah

Solving equations

Solve each equation by the most suitable method. please do not use decimals in the answer.

b)

z(5z-1)=3z(z+3)

c)

x^2+6x-8=0

d)

x^2 +8x/3=1
August 2nd 2006, 11:08 PM
earboth

Quote:

Originally Posted by pashah

Solve each equation by the most suitable method. please do not use decimals in the answer.
b) z(5z-1)=3z(z+3)
c) x^2+6x-8=0
d) x^2 +8x/3=1

Hello, pashah,

to b): rearrange, so that the RHS of the equation is zero. Put the factor z befor the bracket. Then you have two factors. This product can be only zero if one of the factors is zero:
$z(5z-1)-3z(z+3)=0\Longleftrightarrow z(5z-1-3z-9)=0$ $\Longleftrightarrow z(2z-10)=0 \Longleftrightarrow z=0\ \vee\ z=5$

to c) (Are you sure that there isn't a typo? If you have +8 instead of -8 then you could use factorization) Use the formula:
$x_{1,2}=-\frac{6}{2\cdot 1}\pm\sqrt{\frac{6^2-4\cdot 1\cdot (-8)}{4\cdot 1^2}}$ $\ \quad \Longleftrightarrow\ \ x=-3+\sqrt{17}\vee x=-3-\sqrt{17}$

to d) Use the formula:
$x_{1,2}=-\frac{\frac{8}{3}}{2}\pm\sqrt{\frac{\frac{64}{9}-4\cdot 1\cdot (-1)}{4\cdot 1^2}}$ $\Longleftrightarrow$ $x=-\frac{4}{3}+\frac{5}{3}=\frac{1}3}\vee x=-\frac{4}{3}-\frac{5}{3}=-3$

Greetings

EB
August 3rd 2006, 04:37 AM
Soroban

Hello, pashah!

I assume you know how to solve a quadratic equation.
. . Get all terms on the left (0 on the right).
. . Factor the quadratic (if possible).
. . Set each factor equal to zero and solve the resulting equations.

Quote:

$b)\;\;z(5z-1)\:=\:3z(z+3)$

Distribute: . $5z^2 - z\;=\;3x^2 + 9z$

We have: . $2z^2 - 10z\;=\;0$

Factor: . $2z(z - 5)\;=\;0$

And we have: . $\begin{array}{cc}2z\:=\:0\quad\Rightarrow\quad \boxed{z = 0} \\ z - 5\:=\:0\quad\Rightarrow\quad \boxed{z = 5}\end{array}$

Quote:

$c)\;\;x^2 + 6x - 8\:=\:0$

I agree with EB . . . I'm certain there's a typo.

Suppose it was: . $x^2 + 6x$ + $8 \:=\:0$

Factor: . $(x + 2)(x + 4)\:=\:0$

Then we have: . $\begin{array}{cc}x+2\,=\,0\quad\Rightarrow\quad \boxed{x = -2}\\x + 4 \,=\,0\quad\Rightarrow\quad \boxed{x = -4}\end{array}$

Quote:

$d)\;\;x^2 + \frac{8x}{3} \:=\:1$

Multiply by 3: . $3x^2 + 8x\:=\:3$

We have: . $3x^2 + 8x - 3 \:= \:0$

Factor: . $(3x - 1)(x + 3)\:=\:0$

Then we have: . $\begin{array}{cc}3x - 1\:=\:0\quad\Rightarrow\quad \boxed{x = \frac{1}{3}} \\ x + 3\;=\:0\quad\Rightarrow\quad \boxed{x = -3}\end{array}$