Question
Solve the cube root equation 2x2−−−√3=3−5x−−−−−−√3
2
𝑥
2
3
=
3
−
5
𝑥
3
.(1 point)
Responses
x=−52
𝑥
=
−
5
2
and x=1
𝑥
=
1
x equals negative Start Fraction 5 over 2 End Fraction and x equals 1
x=52
𝑥
=
5
2
and x=1
𝑥
=
1
x equals Start Fraction 5 over 2 End Fraction and x equals 1
x=1
𝑥
=
1
and x=32
𝑥
=
3
2
x equals 1 and x equals Start Fraction 3 over 2 End Fraction
x=−3
𝑥
=
−
3
and x=12
𝑥
=
1
2
x equals negative 3 and x equals Start Fraction 1 over 2 End Fraction
2
𝑥
2
3
=
3
−
5
𝑥
3
.(1 point)
Responses
x=−52
𝑥
=
−
5
2
and x=1
𝑥
=
1
x equals negative Start Fraction 5 over 2 End Fraction and x equals 1
x=52
𝑥
=
5
2
and x=1
𝑥
=
1
x equals Start Fraction 5 over 2 End Fraction and x equals 1
x=1
𝑥
=
1
and x=32
𝑥
=
3
2
x equals 1 and x equals Start Fraction 3 over 2 End Fraction
x=−3
𝑥
=
−
3
and x=12
𝑥
=
1
2
x equals negative 3 and x equals Start Fraction 1 over 2 End Fraction
Answers
GPT-4o mini
To solve the equation
\[
2\sqrt[3]{x^2} = 3 - 5\sqrt[3]{x}
\]
we'll start by letting \( y = \sqrt[3]{x} \). Then, we have:
\[
\sqrt[3]{x^2} = y^2 \quad \text{and} \quad \sqrt[3]{x} = y.
\]
Substituting these into the equation gives:
\[
2y^2 = 3 - 5y.
\]
Rearranging this equation leads to:
\[
2y^2 + 5y - 3 = 0.
\]
Next, we can use the quadratic formula to solve for \( y \):
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 2 \), \( b = 5 \), and \( c = -3 \). This gives:
\[
y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}.
\]
Now, we calculate the two possible solutions for \( y \):
1. \( y = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \)
2. \( y = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \)
Now we will convert \( y \) back to \( x \):
1. For \( y = \frac{1}{2} \):
\[
\sqrt[3]{x} = \frac{1}{2} \Rightarrow x = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.
\]
2. For \( y = -3 \):
\[
\sqrt[3]{x} = -3 \Rightarrow x = (-3)^3 = -27.
\]
Thus, the solutions to the original equation are \( x = \frac{1}{8} \) and \( x = -27 \).
Based on the responses given in the options, none of them correspond with our derived values of \( x \). However, they may have been misquoted or incorrect.
If you still wish to solve further for or from any existing options, please clarify or provide more specific contexts around the selections!
\[
2\sqrt[3]{x^2} = 3 - 5\sqrt[3]{x}
\]
we'll start by letting \( y = \sqrt[3]{x} \). Then, we have:
\[
\sqrt[3]{x^2} = y^2 \quad \text{and} \quad \sqrt[3]{x} = y.
\]
Substituting these into the equation gives:
\[
2y^2 = 3 - 5y.
\]
Rearranging this equation leads to:
\[
2y^2 + 5y - 3 = 0.
\]
Next, we can use the quadratic formula to solve for \( y \):
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 2 \), \( b = 5 \), and \( c = -3 \). This gives:
\[
y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-5 \pm \sqrt{25 + 24}}{4} = \frac{-5 \pm \sqrt{49}}{4} = \frac{-5 \pm 7}{4}.
\]
Now, we calculate the two possible solutions for \( y \):
1. \( y = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \)
2. \( y = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \)
Now we will convert \( y \) back to \( x \):
1. For \( y = \frac{1}{2} \):
\[
\sqrt[3]{x} = \frac{1}{2} \Rightarrow x = \left(\frac{1}{2}\right)^3 = \frac{1}{8}.
\]
2. For \( y = -3 \):
\[
\sqrt[3]{x} = -3 \Rightarrow x = (-3)^3 = -27.
\]
Thus, the solutions to the original equation are \( x = \frac{1}{8} \) and \( x = -27 \).
Based on the responses given in the options, none of them correspond with our derived values of \( x \). However, they may have been misquoted or incorrect.
If you still wish to solve further for or from any existing options, please clarify or provide more specific contexts around the selections!