Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive.
Q2= Solve the simultaneous equations for x,y > 0
2log y = log 2 + log x
2^y = 4^x
cheers
for Q1 check the discriminant of the quadratic $\displaystyle b^2 -4ac$
for Q2 using the first equation you should be able to get $\displaystyle \log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.
log both sides on the second equation and it should be easy form there.
An alternative "looking" way to show Q1 would be completing squares and examining the expression.
$\displaystyle
2(x^2 + 3x + \frac92) = 2([x^2 + 2.\frac32.x + (\frac32)^2] + \frac92 - (\frac32)^2) = 2(x + \frac32)^2 + \frac94$
Now since $\displaystyle (x + \frac32)^2$ is always greater than or equal to zero, for all real x, and since $\displaystyle \frac94$ is positive , its sum is strictly positive.
Hello, sparky69er!
1) Show that if $\displaystyle x$ is real, $\displaystyle 2x^2 + 6x + 9$ is always positive.
We have: .$\displaystyle x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$
The sum of two squares is always nonnegative.
When $\displaystyle x = 0$, the polynomial has a minimum value of 9.
2) Solve the simultaneous equations for x, y > 0
. . $\displaystyle \begin{array}{cc}2\log y \:=\: \log 2 + \log x & {\color{blue}[1]} \\ 2^y \: =\: 4^x & {\color{blue}[2]} \end{array}$
Equation [1] is: .$\displaystyle 2\log y \:=\:\log(2x)\;\;{\color{blue}[3]}$
From Equation [2]: .$\displaystyle 2^y \:=\:(2^2)^x \:=\:2^{2x}\quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[4]}$
Substitute into [3]: .$\displaystyle 2\log(2x) \:=\:\log(2x)\quad\Rightarrow\quad\log(2x) \:=\:0$
. . Hence: .$\displaystyle 2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}$
Substitute into [4]: .$\displaystyle y \:=\:2\left(\frac{1}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:1}$
Given a quadratic polynomial equation of the form
$\displaystyle ax^2 + bx + c = 0$,
we have solutions
$\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
(This is the "quadratic formula.")
The "discriminant" is defined as $\displaystyle D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant:
$\displaystyle D > 0 \implies$ two real, unequal roots.
$\displaystyle D = 0 \implies$ one real root. (Or two equal real roots, however you wish to look at it.)
$\displaystyle D < 0 \implies$ two complex conjugate roots.
In the case of D < 0 the curve $\displaystyle y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive.
-Dan