Solving Quadratic Equations and Evaluating Expressions

We are given a quadratic equation \(2x^2 - 7x + 5 = 0\) and asked to find the value of the expression \(x^3 + \frac{125}{8x^3}\).

First, let's solve the given quadratic equation for \(x\). We can factor the equation:

\(2x^2 - 7x + 5 = 0\)

We look for two numbers that multiply to \(2 \times 5 = 10\) and add up to \(-7\). These numbers are \(-2\) and \(-5\).

Rewrite the middle term:

\(2x^2 - 2x - 5x + 5 = 0\)

Group terms and factor:

\(2x(x - 1) - 5(x - 1) = 0\)

\((2x - 5)(x - 1) = 0\)

This gives two possible values for \(x\):

• \(2x - 5 = 0 \implies 2x = 5 \implies x = \frac{5}{2}\)
• \(x - 1 = 0 \implies x = 1\)

Now, let's analyze the expression we need to evaluate: \(x^3 + \frac{125}{8x^3}\). This expression can be rewritten as \(x^3 + \left(\frac{5}{2x}\right)^3\).

This is in the form \(a^3 + b^3\), where \(a = x\) and \(b = \frac{5}{2x}\).

We can use the algebraic identity \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\).

Let's find the value of \(a+b = x + \frac{5}{2x}\) from the original quadratic equation.

Given \(2x^2 - 7x + 5 = 0\). Since we found that \(x=1\) or \(x=5/2\), \(x\) cannot be zero. So, we can divide the entire equation by \(2x\):

\(\frac{2x^2}{2x} - \frac{7x}{2x} + \frac{5}{2x} = \frac{0}{2x}\)

\(x - \frac{7}{2} + \frac{5}{2x} = 0\)

Rearranging the terms to get \(x + \frac{5}{2x}\):

\(x + \frac{5}{2x} = \frac{7}{2}\)

So, we have \(a+b = \frac{7}{2}\).

Now, let's find the value of \(ab\):

\(ab = x \cdot \left(\frac{5}{2x}\right) = \frac{5x}{2x} = \frac{5}{2}\)

Now substitute the values of \(a+b\) and \(ab\) into the identity \(a^3 + b^3 = (a+b)^3 - 3ab(a+b)\):

\(x^3 + \left(\frac{5}{2x}\right)^3 = \left(x + \frac{5}{2x}\right)^3 - 3 \left(x \cdot \frac{5}{2x}\right) \left(x + \frac{5}{2x}\right)\)

\(= \left(\frac{7}{2}\right)^3 - 3 \left(\frac{5}{2}\right) \left(\frac{7}{2}\right)\)

\(= \frac{7^3}{2^3} - \frac{3 \times 5 \times 7}{2 \times 2}\)

\(= \frac{343}{8} - \frac{105}{4}\)

To subtract the fractions, find a common denominator, which is 8:

\(= \frac{343}{8} - \frac{105 \times 2}{4 \times 2}\)

\(= \frac{343}{8} - \frac{210}{8}\)

\(= \frac{343 - 210}{8}\)

\(= \frac{133}{8}\)

The value is \(\frac{133}{8}\). Let's convert this improper fraction to a mixed number. Divide 133 by 8:

\(133 \div 8\). \(133 = 16 \times 8 + 5\).

So, \(\frac{133}{8} = 16\frac{5}{8}\).

Thus, the value of \(x^3 + \frac{125}{8x^3}\) is \(16\frac{5}{8}\).

We can quickly check this for the specific roots found earlier:

• If \(x=1\), \(1^3 + \frac{125}{8(1)^3} = 1 + \frac{125}{8} = \frac{8+125}{8} = \frac{133}{8} = 16\frac{5}{8}\).
• If \(x=\frac{5}{2}\), \(x^3 = \left(\frac{5}{2}\right)^3 = \frac{125}{8}\). \(\frac{125}{8x^3} = \frac{125}{8(125/8)} = \frac{125}{125} = 1\). So \(x^3 + \frac{125}{8x^3} = \frac{125}{8} + 1 = \frac{125+8}{8} = \frac{133}{8} = 16\frac{5}{8}\).

Both roots yield the same result, confirming our general approach using the identity.

Quadratic Equation and Expression Value Revision

Additional Information on Algebraic Manipulation

When dealing with expressions involving powers like \(x^n + \frac{1}{x^n}\) or similar forms derived from a quadratic equation \(ax^2 + bx + c = 0\), a common strategy is to divide the quadratic equation by a power of \(x\) (usually \(x\)) to obtain a relationship between \(x\) and \(\frac{1}{x}\) or related terms like \(\frac{1}{2x}\) in this case.

For \(ax^2 + bx + c = 0\), dividing by \(x\) gives \(ax + b + \frac{c}{x} = 0\), which can be rearranged to get a relationship like \(ax + \frac{c}{x} = -b\). If \(a\) and \(c\) are related (e.g., \(a=1, c=1\) gives \(x+\frac{1}{x}=-b\), or \(a=2, c=5\) gives \(2x + \frac{5}{x} = 7\)), this relationship \(x + k/x = m\) (where \(k\) and \(m\) are constants) is key.

Then, higher powers can be found using identities:

• For \(x^2 + \frac{k^2}{x^2}\): \(\left(x + \frac{k}{x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{k}{x} + \frac{k^2}{x^2} = x^2 + 2k + \frac{k^2}{x^2}\). So, \(x^2 + \frac{k^2}{x^2} = \left(x + \frac{k}{x}\right)^2 - 2k\).
• For \(x^3 + \frac{k^3}{x^3}\): \(\left(x + \frac{k}{x}\right)^3 = x^3 + 3 \cdot x^2 \cdot \frac{k}{x} + 3 \cdot x \cdot \left(\frac{k}{x}\right)^2 + \left(\frac{k}{x}\right)^3 = x^3 + 3kx + \frac{3k^2}{x} + \frac{k^3}{x^3} = \left(x^3 + \frac{k^3}{x^3}\right) + 3k \left(x + \frac{k}{x}\right)\). So, \(x^3 + \frac{k^3}{x^3} = \left(x + \frac{k}{x}\right)^3 - 3k \left(x + \frac{k}{x}\right)\).

In this problem, we had \(x^3 + \left(\frac{5}{2x}\right)^3\), which is of the form \(a^3+b^3\) with \(a=x\) and \(b=\frac{5}{2x}\). The derived relationship was \(x + \frac{5}{2x} = \frac{7}{2}\). Here, the 'k' from the general form \(x+k/x\) is not directly applicable because the coefficient of \(1/x\) is \(5/2\), not a simple constant related to the leading coefficient. However, using the identity \(a^3+b^3\) with \(a=x\) and \(b=\frac{5}{2x}\) correctly utilizes the derived sum \(x + \frac{5}{2x}\) and the product \(x \cdot \frac{5}{2x} = \frac{5}{2}\).