Let's solve your equation step-by-step.
−3x2−4x−4=0
Step 1: Use quadratic formula with a=-3, b=-4, c=-4.
x=
−b±√b2−4ac
2a
x=
−(−4)±√(−4)2−4(−3)(−4)
2(−3)
x=
4±√−32
−6

Answer:
No real solutions.

5 0

3 years ago

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What is the true solution to the equation below? In e^in x + in e^in x^2 = 2 in 8

mariarad [96]

To solve the given equation, w need to review some rules:
$(1) \ ln \ e = 1 \\ 
(2) \ ln \ b^{a} = a*ln \ b \\
(3) \ ln \ a + ln \ b = ln \ (ab) \\
(4) \ ln \ a - ln \ b = ln \ \frac{a}{b} \\$

The given equation is :
$ln \ e^{ln \ x} + ln \ e^{ln \ x^{2}} = 2 \ ln \ 8$
$ln \ x * ln \ e + ln \ x^{2} * ln \ e = 2 \ ln \ 8$ ⇒⇒⇒⇒ rule (2)
$ln \ x + ln \ x^{2} = ln \ 8^{2}$ ⇒⇒⇒⇒ rule (1) and rule (2)

$ln \ ( x* x^{2} ) = ln \ 64$ ⇒⇒⇒⇒ rule (3)
removing the nature logarithm from both sides

$x^{3} = 64 = 4^{3}$
∴ x = 4

So, the correct answer is option (2) ⇒⇒⇒⇒ x = 4

5 0

2 years ago

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