Question

The number of real values of \(\mathrm{x}\) satisfying the equation \(3\left[x^{2}+\left(1 / x^{2}\right)\right]-16[x+(1 / x)]+26=0\) is (a) 1 (b) 2 (c) 3 (d) 4

Short answer

There are 3 real values of \(x\) satisfying the given equation: \(x_1 = 1\), \(x_2 = -1\), and \(x_3 = -\frac{1}{2}\). Therefore, the answer is (c) 3.

Step-by-step solution

Simplify the given equation

Let's first expand the equation and combine like terms. Consider the variables: \(y = x + \frac{1}{x}\), and \(z = x^2 + \frac{1}{x^2}\) Now, the given equation becomes: \(3z -16y + 26=0\)

Relationship between y and z

Take the square of both sides of \(y\) : \(y^2 = x^2 + 2 + \frac{1}{x^2}\) or, \(z = y^2 - 2\), as \(z = x^2 + \frac{1}{x^2}\) Now, we can replace z in our simplified equation.

Solve for y

Replace z with the expression in terms of y : \(3\left(y^2 - 2\right) - 16y + 26 = 0\) or, \(3y^2 - 16y - 2 = 0\) This is a quadratic equation in y. Now, let's find the roots of the quadratic equation.

Find the roots of the quadratic equation

To find the roots, we use the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where a = 3, b = -16, c = -2 Calculating the 2 roots we get: \(y_1 = 2\), and \(y_2 = -\frac{1}{3}\)

Solve for x

Now, substitute the values of y back in the equation \(y = x + \frac{1}{x}\) For \(y_1 = 2\) : \(2 = x + \frac{1}{x}\) \(x^2 - 2x + 1 = 0\) Solving the quadratic equation, we get 1 real value of x : \(x_1 = 1\) For \(y_2 = -\frac{1}{3}\) : \(-\frac{1}{3} = x + \frac{1}{x}\) \(3x^2 + x + 1 = 0\) Solving the quadratic equation, we get 2 real values of x : \(x_2 = -1\), and \(x_3 = -\frac{1}{2}\)

Count the real values of x

We have found 3 distinct real values for x : \(x_1 = 1\), \(x_2 = -1\), and \(x_3 = -\frac{1}{2}\) Therefore, there are 3 real values of \(x\) satisfying the equation, which corresponds to the answer (c) 3.

Key concepts

Solving Quadratic Equations

Quadratic equations are fundamental in algebra, representing curves described by the general form:
\( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is non-zero. Solving these equations means finding the values of the variable \( x \) that make the equation true.

These solutions, called roots, can be real or complex numbers and there can be up to two distinct real roots.

Finding the Roots

• To solve a quadratic equation, you can factorize, complete the square, or use the quadratic formula:
• The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), which gives two solutions for \( x \).
• The term under the square root, \( b^2 - 4ac \), is known as the discriminant and determines the nature of the roots.

If the discriminant is positive, there are two real and distinct roots; if it's zero, there's just one real root (also called a repeated root); and if it's negative, there are two complex roots.

In our exercise, we saw a clever use of substitution to turn the given equation into a quadratic form. By defining \( y = x + \frac{1}{x} \), we could establish a relationship between \( y \) and \( z \), leading us to derive a solvable quadratic equation.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations that define a particular mathematical relationship. In the context of our exercise, the initial expression is complex, involving both \( x \) and its reciprocal.

To simplify and solve complex algebraic expressions, strategical manipulation is often necessary.

Key Strategies for Manipulation:

• Look for patterns or structures in the expression that can be simplified, such as common factors or squares.
• Introduce substitutions to reduce the complexity of the expression, like \( y = x + \frac{1}{x} \) as done in the solution provided.
• Re-express parts of the equation in terms of the substitution to convert it into a familiar form, which is usually easier to solve.

Through these steps, complex algebraic expressions become more manageable, enabling us to use well-known tactics, like solving quadratic equations, to find the solution.

Real Values of a Variable

When we talk about the 'real values of a variable', we are referring to the solutions of an equation that are real numbers. Real numbers include all the numbers on the number line, including positive and negative numbers, zero, and fractions.

Identifying the number of real values a variable can take in an equation is vital, as it informs us about the nature of the equation and can guide us in graphically interpreting the solutions.

Considering the Real Roots

• For quadratic equations, the number of real roots relies heavily on the discriminant \( b^2 - 4ac \).
• In our exercise, we transformed the original equation into a classical quadratic form to determine the real values of \( x \).
• After using the quadratic formula, we concluded that there were distinct real values corresponding to each root of the transposed quadratic equation.

Our approach not only yielded the real roots but also highlighted the relationship between algebraic manipulation and understanding the implications for the values variables can assume in equations.