# Thread: Solving for roots and coefficient

1. ## Solving for roots and coefficient

Solve for the roots and the coefficient k algebraically.
a)The roots of f(x)=x^4 - kx^2 + 9=0 are in arithmetic progression.

And a second problem (same kind)
b) The roots of x^3 - 6x^2 + kx + 64=0 are in geometric progression.

Thanks for the help. ^^

2. Originally Posted by Jukixel
Solve for the roots and the coefficient k algebraically.
a)The roots of f(x)=x^4 - kx^2 + 9=0 are in arithmetic progression.
This can have 0, 2 or 4 real roots, for them to be arithmetic progression there must be 4.

put $y=x^3$, then we have:

$y^2-ky+9=0$

and for the original equation to have four real roots this must have two real roots. These are:

$
x=\pm \sqrt{\left(
\frac{k\pm\sqrt{k^2-36}}{2}
\right)}
$

for these to be real and distinct we require that $k>6$.

For these to be in arithmetic progression there must be an a such that these roots are $-3a,-a,a,3a$

so:

$
3\ \sqrt{\left(
\frac{k -\sqrt{k^2-36}}{2}
\right)}=\sqrt{\left(
\frac{k +\sqrt{k^2-36}}{2}
\right)}
$

Alternative method:

The roots must be $a,\ a+b,\ a+2b,\ a+3b$, for some real $a$ and $b$ so:

$x^4 - kx^2 + 9=(x-a)(x-(a+b))(x-(a+2b))(x-(a+3b))$

Now expand the product on the right, and equate the coefficients of like powers on the left and right to solve for $k,\ a,\ b$

.

3. Sorry, I don't really get the last step. =/

4. Originally Posted by Jukixel
And a second problem (same kind)
b) The roots of x^3 - 6x^2 + kx + 64=0 are in geometric progression.

Thanks for the help. ^^
Here the roots are of the form $a,\ ra,\ r^2a$ for some real $a$ and $r$.

So:

$x^3 - 6x^2 + kx + 64=(x-a)(x-ra)(x-r^2a)$

Now expand the product on the right and equate coefficients of like powers of $x$ to solve for $k,\ a,\ r$

.

5. Oh, I think I get this method. ^^ Thanks a lot!

6. a) Let $x_1=a-3r, \ x_2=a-r, \ x_3=a+r, \ x_4=a+3r$
Then $x_1+x_2+x_3+x_4=0\Rightarrow 4a=0\Rightarrow a=0$
$x_1x_2x_3x_4=9\Rightarrow 9a^4=9\Rightarrow a=\pm 1$

Now find k

7. b) Let $x_1=\frac{a}{q}, \ x_2=a, \ x_3=aq$
Then $x_1x_2x_3=-64\Rightarrow a^3=-64\Rightarrow a=-4$
$x_1+x_2+x_3=6\Rightarrow -\frac{4}{q}-4-4q=6$
Now solve for q, then find k.

8. Hello, Jukixel!

Solve for the roots and the coefficient $k$ algebraically.

a) The roots of $f(x)\:=\:x^4 - kx^2 + 9\:=\:0$ are in arithmetic progression.
The roots are: . $a,\:a+d,\:a+2d,\:a+3d$

The function is: . $f(x) \:=\:x^4 + 0\!\cdot\!x^3 - kx^2 + 0\!\cdot\!x + 9 \:=\:0$

From Vieta's formulas . . .

. . $a + (a+d) + (a+2d)+(a+3d) \:=\:0 \quad\Rightarrow\quad d \:=\:\text{-}\tfrac{2}{3}a$ .[1]

. . $a(a+d)(a+2d)(a+3d) \:=\:9$ .[2]

Substitue [1] into [2]: . $a\left(a-\tfrac{2}{3}a\right)\left(a - \tfrac{4}{3}a\right)(a-2a) \:=\:9$

. . $a\left(\tfrac{1}{3}a\right)\left(\text{-}\tfrac{1}{3}a\right)\left(\text{-}a\right) \:=\:9 \quad\Rightarrow\quad \tfrac{1}{9}a^4 \:=\:9\quad\Rightarrow\quad a^4 \:=\:81 \quad\Rightarrow\quad a \:=\:\pm3$

Substitute into [1]: . $d \:=\:-\tfrac{2}{3}\left(\pm3\right) \:=\:\mp2$

. . Therefore, the roots are: . $\boxed{\text{-}3,\:\text{-}1,\:1,\:3}$

And the polynomial is: . $f(x) \:=\:(x+3)(x+1)(x-1)(x-3) \:=\:x^4 - 10x^2 + 9$

. . Therefore: . $\boxed{k \:=\:10}$

9. Thank you so much!