[SOLVED] equation help!!!!

**liltiger22** · October 5th 2008, 07:34 PM

im in a review section in my class. this is some stuff i havent done in yrs. so if you could help with these 2 problems thatd be great (if possible, can you explain how you got the answer? i really dont remember this from that far back). . .

1. Find the value of K such that the quadratic equation x^2+(2K+1)x-(K/4) = 0 has only one solution

2.Determine the value(s) of x that makes the following statement true: 6x^(4/3) + 7x(2/3) = 3

plz & thank you

**Chris L T521** · October 5th 2008, 07:41 PM

Originally Posted by liltiger22

im in a review section in my class. this is some stuff i havent done in yrs. so if you could help with these 2 problems thatd be great (if possible, can you explain how you got the answer? i really dont remember this from that far back). . .

1. Find the value of K such that the quadratic equation x^2+(2K+1)x-(K/4) = 0 has only one solution

A quadratic equation only has one solution when the discriminant is equal to zero.

Note that the discriminant is defined by $\Delta=\sqrt{b^2-4ac}$

Since our quadratic equation has the form $ax^2+bx+c=0$ , we see that the discriminant then would be $\sqrt{(2k+1)^2+4(1)(K/4)}$

Set this equal to zero and solve for K.

Can you try to take it from here?

2.Determine the value(s) of x that makes the following statement true: 6x^(4/3) + 7x(2/3) = 3

plz & thank you

With a little rearrangement, we see that this equation can be written as $6(x^{\frac{2}{3}})^2+7x^{\frac{2}{3}}=3$

Now make a little substitution: let's say $z=x^{\frac{2}{3}}$

Then this becomes a quadratic equation: $6z^2+7z=3\implies 6z^2+7z-3=0$

Solve this equation, and then back substitute $z=x^{\frac{2}{3}}$ to solve for x.

Can you try to take it from here?

Does this make sense?

--Chris

**icemanfan** · October 5th 2008, 07:43 PM

1. In order for the quadratic equation $ax^2 + bx + c = 0$ to have only one solution, the discriminant $b^2 - 4ac$ must be equal to zero.

The discriminant for $x^2 + (2k+1)x - k/4 = 0$ is $(2k+1)^2 - 4(1)(-k/4) = (2k + 1)^2 + k$ .

Hence $(2k + 1)^2 + k = 0$

$4k^2 + 4k + 1 + k = 0$

$4k^2 + 5k + 1 = 0$

$(4k + 1)(k + 1) = 0$

so k = -1 or k = -1/4.

**earboth** · October 5th 2008, 07:50 PM

Originally Posted by icemanfan

1. In order for the quadratic equation $ax^2 + bx + c = 0$ to have only one solution, the discriminant $b^2 - 4ac$ must be equal to zero.

The discriminant for $x^2 + (2k+1)x - k/4 = 0$ is $(2k+1)^2 - 4(1)(k/4) = (2k + 1)^2 - k$ .

Hence $(2k + 1)^2 - k = 0$

$4k^2 + 4k + 1 - k = 0$

$4k^2 + 3k + 1 = 0$

$k = \frac{-3 \pm \sqrt{3^2 - 4(4)(1)}}{8}$

so there are no real solutions.

I don't want to pick at you but shouldn't it be:

$x^2 + (2k+1)x - k/4 = 0$ is $(2k+1)^2 - 4(1)(\bold{{\color{red}-}}k/4) = (2k + 1)^2\bold{{\color{red}+}} k$ .

**icemanfan** · October 5th 2008, 07:52 PM

Originally Posted by earboth

I don't want to pick at you but shouldn't it be:

$x^2 + (2k+1)x - k/4 = 0$ is $(2k+1)^2 - 4(1)(\bold{{\color{red}-}}k/4) = (2k + 1)^2\bold{{\color{red}+}} k$ .

You're right. I'm going to edit my post now.

**liltiger22** · October 5th 2008, 08:11 PM

well i got number 2, but im still workin on number 1 . . . but if you could help on number 1, thatd be great

Thread: [SOLVED] equation help!!!!

LinkBack

Thread Tools

Display

[SOLVED] equation help!!!!

Similar Math Help Forum Discussions

[SOLVED] Equation of a plane - Given z intercept and parametric equation

[SOLVED] equation

[SOLVED] Help w/ an equation please

[SOLVED] Equation Help

[SOLVED] equation help

Search Tags