Solve. 6+2x^(2)-3x=8x^(2) Choose 1 answer: (A) x=3,-(1)/(2) (B) x=(5+-sqrt57)/(16) (c) x=(1+-sqrt17)/(-4) (D) x=(-4+-sqrt34)/(3)

Setting the equation to zero: First, we need to set the equation to zero by moving all terms to one side. 6 + 2x^2 - 3x = 8x^2 Subtract 8x^2 from both sides to get: 6 - 3x - 6x^2 = 0 Rearrange the terms to get a standard quadratic equation form: -6x^2 - 3x + 6 = 0 Now, we can multiply the entire equation by -1 to make the x^2 coefficient positive: 6x^2 + 3x - 6 = 0Rearranging the terms: Now, we will use the quadratic formula to solve for x. The quadratic formula is: x = (-b ± sqrt(b^2 - 4ac)) / (2a) For our equation, a = 6, b = 3, and c = -6.Using the quadratic formula: Let's calculate the discriminant (the part under the square root in the quadratic formula): Discriminant = b^2 - 4ac Discriminant = (3)^2 - 4(6)(-6) Discriminant = 9 + 144 Discriminant = 153Calculating the discriminant: Now we can substitute a, b, and the discriminant into the quadratic formula: x = (-3 ± sqrt(153)) / (2*6) x = (-3 ± sqrt(153)) / 12Substituting values into the quadratic formula: We can simplify sqrt(153) to get the exact solutions: x = (-3 ± sqrt(153)) / 12 Since sqrt(153) cannot be simplified to an integer or a simple fraction, we will leave it as is.Simplifying the square root: Now we have two possible solutions for x: x = (-3 + sqrt(153)) / 12 or x = (-3 - sqrt(153)) / 12 These are the exact solutions in their simplest form.

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6+2x^(2)-3x=8x^(2)
Choose 1 answer:
(A)
x=3,-(1)/(2)
(B)
x=(5+-sqrt57)/(16)
(c)
x=(1+-sqrt17)/(-4)
(D)
x=(-4+-sqrt34)/(3)

Solve. $\newline$ $6+2 x^{2}-3 x=8 x^{2}$ $\newline$ Choose $1$ answer: $\newline$ (A) $x=3,-\frac{1}{2}$ $\newline$ (B) $x=\frac{5 \pm \sqrt{57}}{16}$ $\newline$ (C) $x=\frac{1 \pm \sqrt{17}}{-4}$ $\newline$ (D) $x=\frac{-4 \pm \sqrt{34}}{3}$

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Math Problems

Algebra 2

Solve a quadratic equation using the quadratic formula

Full solution

Q. Solve.

\newline

6+2 x^{2}-3 x=8 x^{2}

\newline

Choose

1

answer:

\newline

(A)

x=3,-\frac{1}{2}

\newline

(B)

x=\frac{5 \pm \sqrt{57}}{16}

\newline

(C)

x=\frac{1 \pm \sqrt{17}}{-4}

\newline

(D)

x=\frac{-4 \pm \sqrt{34}}{3}

Setting the equation to zero: First, we need to set the equation to zero by moving all terms to one side. $\newline$ $6 + 2x^2 - 3x = 8x^2$ $\newline$ Subtract $8x^2$ from both sides to get: $\newline$ $6 - 3x - 6x^2 = 0$ $\newline$ Rearrange the terms to get a standard quadratic equation form: $\newline$ $-6x^2 - 3x + 6 = 0$ $\newline$ Now, we can multiply the entire equation by $-1$ to make the $x^2$ coefficient positive: $\newline$ $6x^2 + 3x - 6 = 0$
Rearranging the terms: Now, we will use the quadratic formula to solve for $x$ . The quadratic formula is: $\newline$ $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$ $\newline$ For our equation, $a = 6$ , $b = 3$ , and $c = -6$ .
Using the quadratic formula: Let's calculate the discriminant (the part under the square root in the quadratic formula): $\newline$ Discriminant = $b^2 - 4ac$ $\newline$ Discriminant = $(3)^2 - 4(6)(-6)$ $\newline$ Discriminant = $9 + 144$ $\newline$ Discriminant = $153$
Calculating the discriminant: Now we can substitute $a$ , $b$ , and the discriminant into the quadratic formula: $\newline$ $x = \frac{{-3 \pm \sqrt{153}}}{{2 \cdot 6}}$ $\newline$ $x = \frac{{-3 \pm \sqrt{153}}}{{12}}$
Substituting values into the quadratic formula: We can simplify $\sqrt{153}$ to get the exact solutions: $\newline$ $x = \frac{-3 \pm \sqrt{153}}{12}$ $\newline$ Since $\sqrt{153}$ cannot be simplified to an integer or a simple fraction, we will leave it as is.
Simplifying the square root: Now we have two possible solutions for x: $\newline$ $x = \frac{{-3 + \sqrt{153}}}{{12}}$ or $x = \frac{{-3 - \sqrt{153}}}{{12}}$ $\newline$ These are the exact solutions in their simplest form.