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Citrus2011 [14]
3 years ago
11

Subtract. Simplify, if possible. 5/8 - * 2 1 / 2 - 3

Mathematics
1 answer:
Nat2105 [25]3 years ago
8 0

2\frac{1}{2} - \frac{5}{8}

Convert the mixed number to an improper fraction:

\frac{5}{2} - \frac{5}{8}

Subtract the fractions to get your final answer:

\frac{15}{8}

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What are the solutions of the equation 9x4 – 2x2 – 7 = 0? Use u substitution to solve. tions of the equation 9
skelet666 [1.2K]

Answer:

Step-by-step explanation:

Let u^2=x^4\\u = x^2

Subbing in:

9u^2-2u-7=0

a = 9, b = -2, c = -7

The product of a and c is the aboslute value of -63, so a*c = 63.  We need 2 factors of 63 that will add to give us -2.  The factors of 63 are {1, 63}, (3, 21}, {7, 9}.  It looks like the combination of -9 and +7 will work because -9 + 7 = -2.  Plug in accordingly:

9u^2-9u+7u-7=0

Group together in groups of 2:

(9u^2-9u)+(7u-7)=0

Now factor out what's common within each set of parenthesis:

9u(u-1)+7(u-1)=0

We know this combination "works" because the terms inside the parenthesis are identical.  We can now factor those out and what's left goes together in another set of parenthesis:

(u-1)(9u+7)=0

Remember that u=x^2

so we sub back in and continue to factor.  This was originally a fourth degree polynomial; that means we have 4 solutions.

(x^2-1)(9x^2+7)=0

The first two solutions are found withing the first set of parenthesis and the second two are found in other set of parenthesis.  Factoring (x^2-1) gives us that x = 1 and -1.  The other set is a bit more tricky.  If

9x^2+7=0 then

9x^2=-7 and

x^2=-\frac{7}{9}

You cannot take the square root of a negative number without allowing for the imaginary component, i, so we do that:

x=±\sqrt{-\frac{7}{9} }

which will simplify down to

x=±\frac{\sqrt{7} }{3}i

Those are the 4 solutions to the quartic equation.

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Step-by-step explanation:

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