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Which shows the correct substitution of the values [tex]\( a, b \)[/tex], and [tex]\( c \)[/tex] from the equation [tex]\( 1 = -2x + 3x^2 + 1 \)[/tex] into the quadratic formula?

Quadratic formula: [tex] x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

A. [tex] x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(1)}}{2(3)} \)

B. [tex] x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(9)(9)}}{2(0)} \)

C. [tex] x = \frac{-(-2) \div \sqrt{(-2)^2 - 4(0)(2)}}{2(0)} \)

D. [tex] x = \frac{-11 \sqrt{3^2 - 4(-2)(0)}}{2(-2)} \)


Sagot :

Let's start by looking at the equation given: [tex]\(1 = -2x + 3x^2 + 1\)[/tex]. We can rewrite this equation in the standard quadratic form [tex]\(ax^2 + bx + c = 0\)[/tex].

First, subtract 1 from both sides of the equation:
[tex]\[1 - 1 = -2x + 3x^2 + 1 - 1\][/tex]
[tex]\[0 = -2x + 3x^2\][/tex]

Which simplifies to:
[tex]\[3x^2 - 2x = 0\][/tex]

Here, the quadratic equation we have is:
[tex]\[3x^2 - 2x = 0\][/tex]

In the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]:
[tex]\[a = 3\][/tex]
[tex]\[b = -2\][/tex]
[tex]\[c = 0\][/tex]

The quadratic formula is:
[tex]\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\][/tex]

Substitute the values [tex]\(a = 3\)[/tex], [tex]\(b = -2\)[/tex], and [tex]\(c = 0\)[/tex] into the quadratic formula:

[tex]\[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(0)}}{2(3)}\][/tex]

Simplify the expression step by step:

1. Calculate the term [tex]\(-b\)[/tex]:
[tex]\[ -(-2) = 2\][/tex]

2. Calculate the term [tex]\(b^2\)[/tex]:
[tex]\[(-2)^2 = 4\][/tex]

3. Calculate the discriminant ([tex]\(\Delta\)[/tex]), which is the term [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[\Delta = 4 - 4(3)(0)\][/tex]
[tex]\[\Delta = 4 - 0\][/tex]
[tex]\[\Delta = 4\][/tex]

So, the correct substitution into the quadratic formula is:
[tex]\[x = \frac{2 \pm \sqrt{4}}{6}\][/tex]
[tex]\[x = \frac{2 \pm 2}{6}\][/tex]

Thus, the correct substitution of the values [tex]\(a, b\)[/tex], and [tex]\(c\)[/tex] from the equation into the quadratic formula is:
[tex]\[x = \frac{2 \pm \sqrt{4 - 12}}{6}\][/tex]
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