1. ## Algebra

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

2. for Q1 check the discriminant of the quadratic $b^2 -4ac$

for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

log both sides on the second equation and it should be easy form there.

3. Originally Posted by sparky69er
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
An alternative "looking" way to show Q1 would be completing squares and examining the expression.
$
2(x^2 + 3x + \frac92) = 2([x^2 + 2.\frac32.x + (\frac32)^2] + \frac92 - (\frac32)^2) = 2(x + \frac32)^2 + \frac94$

Now since $(x + \frac32)^2$ is always greater than or equal to zero, for all real x, and since $\frac94$ is positive , its sum is strictly positive.

4. Originally Posted by bobak
for Q1 check the discriminant of the quadratic $b^2 -4ac$

for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

log both sides on the second equation and it should be easy form there.
Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????

5. Hello, sparky69er!

1) Show that if $x$ is real, $2x^2 + 6x + 9$ is always positive.

We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

The sum of two squares is always nonnegative.

When $x = 0$, the polynomial has a minimum value of 9.

2) Solve the simultaneous equations for x, y > 0

. . $\begin{array}{cc}2\log y \:=\: \log 2 + \log x & {\color{blue}[1]} \\ 2^y \: =\: 4^x & {\color{blue}[2]} \end{array}$

Equation [1] is: . $2\log y \:=\:\log(2x)\;\;{\color{blue}[3]}$

From Equation [2]: . $2^y \:=\:(2^2)^x \:=\:2^{2x}\quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[4]}$

Substitute into [3]: . $2\log(2x) \:=\:\log(2x)\quad\Rightarrow\quad\log(2x) \:=\:0$

. . Hence: . $2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}$

Substitute into [4]: . $y \:=\:2\left(\frac{1}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:1}$

6. Originally Posted by sparky69er
Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????
Given a quadratic polynomial equation of the form
$ax^2 + bx + c = 0$,
we have solutions
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The "discriminant" is defined as $D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant:
$D > 0 \implies$ two real, unequal roots.

$D = 0 \implies$ one real root. (Or two equal real roots, however you wish to look at it.)

$D < 0 \implies$ two complex conjugate roots.

In the case of D < 0 the curve $y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive.

-Dan

7. Originally Posted by Soroban
Hello, sparky69er!

We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

The sum of two squares is always nonnegative.
Smart, really smart solution