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Dimas [21]
3 years ago
8

Find all real solutions of the equation. x4/3 − 7x2/3 + 10 = 0

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
8 0
Solve by Factoring x^(2/3)-7x^(1/3)+10=0
x
2
3
−
7
x
1
3
+
10
=
0
x
2
3
-
7
x
1
3
+
10
=
0
Rewrite
x
2
3
x
2
3
as
(
x
1
3
)
2
(
x
1
3
)
2
.
(
x
1
3
)
2
−
7
x
1
3
+
10
=
0
(
x
1
3
)
2
-
7
x
1
3
+
10
=
0
Let
u
=
x
1
3
u
=
x
1
3
. Substitute
u
u
for all occurrences of
x
1
3
x
1
3
.
u
2
−
7
u
+
10
=
0
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Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse
Drupady [299]

Answer:

a'(d) = \frac{d}{5} + \frac{3}{5}

a(a'(d)) = a'(a(d)) = d

Step-by-step explanation:

Given

a(d) = 5d - 3

Solving (a): Write as inverse function

a(d) = 5d - 3

Represent a(d) as y

y = 5d - 3

Swap positions of d and y

d = 5y - 3

Make y the subject

5y = d + 3

y = \frac{d}{5} + \frac{3}{5}

Replace y with a'(d)

a'(d) = \frac{d}{5} + \frac{3}{5}

Prove that a(d) and a'(d) are inverse functions

a'(d) = \frac{d}{5} + \frac{3}{5} and a(d) = 5d - 3

To do this, we prove that:

a(a'(d)) = a'(a(d)) = d

Solving for a(a'(d))

a(a'(d))  = a(\frac{d}{5} + \frac{3}{5})

Substitute \frac{d}{5} + \frac{3}{5} for d in  a(d) = 5d - 3

a(a'(d))  = 5(\frac{d}{5} + \frac{3}{5}) - 3

a(a'(d))  = \frac{5d}{5} + \frac{15}{5} - 3

a(a'(d))  = d + 3 - 3

a(a'(d))  = d

Solving for: a'(a(d))

a'(a(d)) = a'(5d - 3)

Substitute 5d - 3 for d in a'(d) = \frac{d}{5} + \frac{3}{5}

a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}

Add fractions

a'(a(d)) = \frac{5d - 3+3}{5}

a'(a(d)) = \frac{5d}{5}

a'(a(d)) = d

Hence:

a(a'(d)) = a'(a(d)) = d

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