Mathematics

Question

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

2 Answer

  • Good evening ,

    Answer :

    A.

    x = ±20

    Step-by-step explanation:

    x² − 1 = 399 ⇔ x²= 400 ⇌ x² = 20² ⇌ x² − 20² = 0 ⇌ (x-20)(x+20) = 0

    ⇌ x = 20 or x = -20.

    :)

  • 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}-1-399=0[/tex]

    Subtract the numbers

    [tex]&x^{2}-1-399=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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