Question

For which value of \(b\), equation \(x^{2}+b x-1=0\) and \(\mathrm{x}^{2}+\mathrm{x}+\mathrm{b}=0\) have a common root. (a) \(-\sqrt{2}\) (b) \(-\mathrm{i} \sqrt{3}\) (c) \(\mathrm{i} \sqrt{5}\) (d) \(\sqrt{2}\)

Short answer

The correct value of b for which both equations have a common root is (d) \(\sqrt{2}\).

Step-by-step solution

Assume the common root

Let the common root for both equations be p. Then the equations can be rewritten as follows: 1. \(p^2 + bp -1 = 0\) 2. \(p^2 + p + b = 0\)

Solve for b

Using the two equations we can solve for b in terms of p. From equation 1, solve for b: \(b = \frac{1 - p^2}{p}\)

Substitute in equation 2

Now, substitute the expression for b in the second equation: \(p^2 + p + (\frac{1 - p^2}{p}) = 0\)

Solve the equation for p

Now, we want to find the correct value of p for which there exists a common root: \(p^3 + p^2 + 1 - p^2 = 0\) So, the equation simplifies to: \(p^3 + 1 = 0\) Thus, \(p = -1\)

Find the value of b

Now that we have found the value of p, we can find the value of b using the expression we derived in Step 2: \(b = \frac{1 - (-1)^2}{-1} = -1\)

Check which option gives the correct b value

Plug in the given options for b into the expression for b in terms of p, and see which one gives us b = -1. (a) \(-\sqrt{2}\): \(b = \frac{1-(-\sqrt{2})^2}{-\sqrt{2}} = \frac{3}{\sqrt{2}} \neq -1\) (b) \(-i\sqrt{3}\): \(b = \frac{1-(-i\sqrt{3})^2}{-i\sqrt{3}} = \frac{1 - (-3)}{-i\sqrt{3}} = \frac{4}{-i\sqrt{3}}\) (c) \(i\sqrt{5}\): \(b = \frac{1-(i\sqrt{5})^2}{i\sqrt{5}} = \frac{1 - (-5)}{i\sqrt{5}} = \frac{6}{i\sqrt{5}}\) (d) \(\sqrt{2}\): \(b = \frac{1-(\sqrt{2})^2}{\sqrt{2}} = \frac{-1}{1} = -1\) From the above calculations, we can see that option (d) \(\sqrt{2}\) when substituted in the expression, gives b = -1. Therefore, answer choice (d) is the correct value of b for which both equations have a common root.

Key concepts

Solving Quadratic Equations

When delving into the realm of algebra, one of the fundamental skills is solving quadratic equations. These equations take the form of ax^2 + bx + c = 0, where a, b, and c are constants, and x represents the unknown variable. The solutions to these equations, known as roots, can be found using various methods including factoring, completing the square, using the quadratic formula, or graphing.

To solve a quadratic equation and find the roots, the equation has to be set to zero. This creates a standard form that allows one to apply the above methods to solve for x. Each method has its own place, and the most efficient one often depends on the specific form of the quadratic equation at hand.

For instance, in the textbook exercise provided, the substitution method is used to find a common root between two different quadratic equations. This method cleverly reduces the problem to one variable, streamlining the path to a solution.

System of Equations

Intersecting paths in mathematics can sometimes be represented by a system of equations, a set of two or more equations with a common variable. The overlap between these paths—the points where the equations are both true—is what we seek when solving such a system. There are multiple ways to find the solutions to a system of equations, like graphing, substitution, and elimination.

In graphing, equations are plotted on a coordinate axis, and their points of intersection are identified. Substitution involves solving one equation for a variable and substituting the result into another equation, simplifying the system to one with a single variable. Elimination method focuses on adding or subtracting equations to cancel out one of the variables, again reducing it to a single-variable equation.

Using these methods, we can determine if the system has no solution, one solution, or infinitely many solutions. The exercise in question engages the substitution method within a system of quadratic equations to discover a shared root, illustrating just one way systems of equations can intersect.

Substitution Method

As seen in our exercise, the substitution method is a powerful tool for solving systems of equations. This technique involves rearranging one of the equations to express one variable in terms of the others, then substituting this expression into another equation. This process eliminates one variable, allowing us to solve the system with greater ease.

The substitution method is particularly useful when equations cannot be easily manipulated for elimination or when the system includes nonlinear equations, such as quadratics. To apply this method effectively, one must isolate a variable in one equation and ensure the substitution is clearly followed through when inserted into the other equations of the system.

This method not only simplifies the solving process but also encourages a meticulous and orderly approach to solving complex algebraic problems. As we've observed in the given example, methodical substitution led to the discovery of the common root and the subsequent identification of the correct value of b in the quadratic equation.