Marc solves the equation Latex: (\sqrt[4]{x})^3=5\textsf{.} His work is shown below. Step 1: Latex: (\sqrt[4]{x})^3=5
(
√
[
4
]
x
)
3
=
5

Step 2: Latex: x^{\frac{3}{4} }=5
x
3
4
=
5

Step 3: Latex: x^{\frac{3}{4} \cdot \frac{4}{3} }=5^{\frac{4}{3} }
x
3
4
⋅
4
3
=
5
4
3

Step 4: Latex: x=\sqrt[3]{5 \cdot 5 \cdot 5 \cdot 5}
x
=
√
[
3
]
5
⋅
5
⋅
5
⋅
5

Step 5: Latex: \require{cancel}x=5\sqrt[3]{\cancel {5} \cdot \cancel{5} \cdot \cancel{5} \cdot 5}

Step 6: Latex: x=5\sqrt[3]{5}
x
=
5
√
[
3
]
5

Explain what Marc did in steps 4 and 5.

Why did he do this?

Create your own radical equation and explain how to solve it.

Is there an extraneous solution to your equation?

50% of students voted for candidate b.

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12