MHB Solve exponential equation x^4 = (5x+6)^2

[ketanco]

x^4 = (5x+6)^2

then what is the multiplication product of all values x can take?

i tried taking square roots of each and wrote in absolute value and found 6, 1, -1 (may be wrong) already but there must be more or different because it is not even in answer choices and the answer should be -36

 

[skeeter]

x^4 = (5x+6)^2

then what is the multiplication product of all values x can take?

i tried taking square roots of each and wrote in absolute value and found 6, 1, -1 (may be wrong) already but there must be more or different because it is not even in answer choices and the answer should be -36

$x^4 - (5x+6)^2 = 0$

$[x^2 - (5x+6)] \cdot [x^2 + (5x+6)] = 0$

$[(x-6)(x+1)] \cdot [(x+2)(x+3)] = 0$

$x \in \{-3,-2,-1,6 \}$

 

[ketanco]

great, and thanks

and what if we tried to solve with absolute value like i tried by taking square roots of both sides? can it be done?

if so how?

if not why not?

 

[skeeter]

$\sqrt{x^4} = \sqrt{(5x+6)^2}$

$|x^2| = |5x+6|$note ... $|x^2| = x^2$

$|5x+6| = 5x+6$ if $5x+6 \ge 0$

$|5x+6| = -(5x+6)$ if $5x+6 < 0$case 1

$x^2 = 5x + 6$ if $5x+6 \ge 0 \implies x \ge -\dfrac{6}{5}$

$x^2 - 5x - 6 = 0$

$(x-6)(x+1) = 0$ ... both zeros are $\ge -\dfrac{6}{5}$case 2

$x^2 = -(5x+6)$ if $5x+6 < 0 \implies x < -\dfrac{6}{5}$

$x^2 + 5x + 6 = 0$

$(x+3)(x+2) = 0$ ... both zeros are $< -\dfrac{6}{5}$

 

[RLBrown]

x^4 = (5x+6)^2

then what is the multiplication product of all values x can take?

i tried taking square roots of each and wrote in absolute value and found 6, 1, -1 (may be wrong) already but there must be more or different because it is not even in answer choices and the answer should be -36

xxxx-(5x+6)(5x+6)=0
xxxx-25xx-60x-36=0
(x-a1)(x-a2)(x-a3)(x-a4)=0

a1a2a3a4 = ?

 

[skeeter]

xxxx-(5x+6)(5x+6)=0
xxxx-25xx-60x-36=0
(x-a1)(x-a2)(x-a3)(x-a4)=0

a1a2a3a4 = ?

$x^4 - 25x^2 - 60x - 36 = 0$

try using the rational root theorem ...