Solve the exponential equation $e^x\left(2x^2+5x-7\right)=0$

Step-by-step Solution

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Final answer to the problem

$x=1,\:x=-\frac{7}{2}$
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Step-by-step Solution

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  • Solve by completing the square
  • Solve for x
  • Find the derivative using the definition
  • Solve by quadratic formula (general formula)
  • Simplify
  • Find the integral
  • Find the derivative
  • Factor
  • Factor by completing the square
  • Find the roots
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1

Factor the trinomial $\left(2x^2+5x-7\right)$ of the form $ax^2+bx+c$, first, make the product of $2$ and $-7$

$\left(2\right)\left(-7\right)=-14$

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$\left(2\right)\left(-7\right)=-14$

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Learn how to solve exponential equations problems step by step online. Solve the exponential equation e^x(2x^2+5x+-7)=0. Factor the trinomial \left(2x^2+5x-7\right) of the form ax^2+bx+c, first, make the product of 2 and -7. Now, find two numbers that multiplied give us -14 and add up to 5. Rewrite the original expression. Factor \left(2x^2+7x-2x-7\right) by the greatest common divisor 2.

Final answer to the problem

$x=1,\:x=-\frac{7}{2}$

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Function Plot

Plotting: $e^x\left(2x^2+5x-7\right)$

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Exponential Equations

Exponential equations are those where the unknown appears only in the exponents of powers of constant bases.

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