What are the solutions to the equation x2 − 1 = 399? A. X=20 and x=20 B. X=200 and x=200 C. X=400 and x=400 D. X=398 squared and x=398 squared
Question
A. X=20 and x=20
B. X=200 and x=200
C. X=400 and x=400
D. X=398 squared and x=398 squared
2 Answer

1. User Answers profarouk
Good evening ,
Answer :
A.
x = ±20
Stepbystep explanation:
x² − 1 = 399 ⇔ x²= 400 ⇌ x² = 20² ⇌ x² − 20² = 0 ⇌ (x20)(x+20) = 0
⇌ x = 20 or x = 20.
:)

2. User Answers 88op
X=20 and x=20 are the solutions to the equation [tex]x^{2} 1 = 399[/tex]
What is a Quadratic equation?
A quadratic equation exists as an algebraic equation of the second degree in x. The quadratic equation in its standard form exists [tex]ax^{2} + bx + c = 0[/tex], where a and b exist as the coefficients, x is the variable, and c stands as the constant term.
Given,
[tex]x^{2} 1 = 399[/tex]
To find,
The solutions to the equation.
Step 1
[tex]x^{2} 1 = 399[/tex]
Move terms to the left side
[tex]&x^{2}1=399 \\[/tex]
[tex]&x^{2}1399=0[/tex]
Subtract the numbers
[tex]&x^{2}1399=0 \\[/tex]
[tex]&x^{2}400=0[/tex]
Use the quadratic formula
[tex]$x=\frac{b \pm \sqrt{b^{2}4 a c}}{2 a}$[/tex]
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic equation.
[tex]&x^{2}400=0 \\[/tex]
a=1
b=0
c=400
[tex]&x=\frac{0 \pm \sqrt{0^{2}4 \cdot 1(400)}}{2 \cdot 1}[/tex]
[tex]&x=\frac{0 \pm \sqrt{(0^{}1600)}}{2 }[/tex]
[tex]$x=\frac{\pm 40}{2}$[/tex]
[tex]$x=\frac{40}{2}$[/tex]
[tex]$x=\frac{40}{2}$[/tex]
Hence,
[tex]$x=20$[/tex]
[tex]$x=20$[/tex]
Thus, Option A. X=20 and x=20 are the solutions to the equation [tex]x^{2} 1 = 399[/tex]
To learn more about Quadratic equations refer to:
https://brainly.com/question/1214333
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