Problem 18 - Intermediate Algebra

Question

Solve. \(2 x+\sqrt{x+1}=8\)

Step-by-Step Solution

Answer

The solution is \(x = 3\).

Step 1: Isolate the Square Root

This equation features a square root, so let's begin by isolating it. Start with the original equation:\[2x + \sqrt{x+1} = 8\]Subtract \(2x\) from both sides:\[\sqrt{x+1} = 8 - 2x\]

Step 2: Square Both Sides

Remove the square root by squaring both sides of the equation. This will help eliminate the radical:\[\left(\sqrt{x+1}\right)^2 = (8 - 2x)^2\]This simplifies to:\[x + 1 = (8 - 2x)^2\]

Step 3: Expand the Squared Expression

Expand \((8 - 2x)^2\) using the formula \((a-b)^2 = a^2 - 2ab + b^2\):\[x+1 = 64 - 32x + 4x^2\]

Step 4: Rearrange Into a Quadratic Equation

Bring all terms to one side of the equation to form a quadratic equation:\[4x^2 - 33x + 63 = 0\]

Step 5: Solve the Quadratic Equation

Now we solve the quadratic equation \(4x^2 - 33x + 63 = 0\) using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -33\), and \(c = 63\):Compute the discriminant: \[b^2 - 4ac = (-33)^2 - 4 \times 4 \times 63\]\[= 1089 - 1008 = 81\]Since the discriminant is positive, there are two real solutions:\[x = \frac{33 \pm \sqrt{81}}{8}\]\[x = \frac{33 \pm 9}{8}\]This gives solutions:\[x = \frac{33 + 9}{8} = 5.25,\]\[x = \frac{33 - 9}{8} = 3\]

Step 6: Check Each Solution

Substitute each solution back into the original equation to verify:For \(x = 5.25\):\[2(5.25) + \sqrt{5.25 + 1} = 10.5 + \sqrt{6.25} eq 8\]This solution does not satisfy the original equation.For \(x = 3\):\[2(3) + \sqrt{3 + 1} = 6 + 2 = 8\]This solution satisfies the original equation.

Key Concepts

Understanding Quadratic Equations

Quadratic equations are a fundamental part of algebra, commonly taking the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, with \(a\) not equal to zero, making \(ax^2\) the highest-degree term. Quadratic equations can arise in various real-world scenarios, such as calculating areas, optimizing problems, or modeling trajectories. They are vital for understanding complex algebraic expressions that encapsulate real-world phenomena.

• Standard Form: A quadratic equation should be arranged in the form \(ax^2 + bx + c = 0\) for streamlined solving.
• Importance: Solving these equations allows us to find unknown values represented by \(x\), revealing crucial insights into the problem at hand.

Recognizing and forming quadratic equations, as shown by rearranging terms, is the first step in solving them effectively. This is essential when solving more complex equations, like those involving radicals, by ultimately converting them into a quadratic form.

Discriminant: Key to Solutions

The discriminant of a quadratic equation gives critical information about the nature of its roots without solving the equation directly. It is derived from the formula \(b^2 - 4ac\) from the quadratic equation \(ax^2 + bx + c = 0\).The discriminant helps determine how many and what kind of solutions exist:

• Positive Discriminant: Indicates two distinct real roots.
• Zero Discriminant: Results in a double root, meaning the equation has one real root repeated twice.
• Negative Discriminant: Suggests there are no real roots, and the solutions are complex or imaginary.

In our example, the discriminant was computed as 81, because \(b^2 - 4ac = 1089 - 1008 = 81\). Since it's positive, this assured us of the existence of two distinct real solutions. Understanding the discriminant is crucial when solving quadratic equations, particularly when determining the number of valid solutions.

Importance of Verifying Solutions

Verifying solutions in mathematical equations, particularly when involving radicals and quadratics, is an essential step to ensure accuracy. When an equation undergoes operations like squaring, it can introduce extraneous solutions that don't work in the original equation.To verify, substitute the solution back into the original equation:

• For \(x = 5.25\), substitution shows it doesn't satisfy the original equation.
• Verification for \(x = 3\) confirms it satisfies \(2x + \sqrt{x+1} = 8\).

Checking after solving is significant because it confirms which solutions hold true under the equation's initial conditions. Especially in equations with radicals, like the initial step of isolating and squaring both sides, verification prevents accepting invalid solutions inadvertently introduced.