Question

Solve the quadratic equation using the Quadratic Formula. Then solve the equation using another method. Which method do you prefer? Explain.\(5 x^2+38=3\)

Short answer

Both the Quadratic Formula and Factoring have determined that there are no real roots for the equation \(5x^2+35=0\). The preference for method should be determined by the individual solving it based on comfort and ease with each method.

Step-by-step solution

Re-arrange the equation in standard form

Convert the equation \(5x^2+38=3\) into the standard form \(ax^2+bx+c=0\). By moving the term '3' to the left side of the equation, it becomes: \(5x^2+38-3=0\), which simplifies to \(5x^2+35=0\). Here, \(a=5\), \(b=0\), and \(c=35\).

Apply the Quadratic Formula

The Quadratic Formula is given as \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Substituting \(a=5\), \(b=0\), and \(c=35\) into the formula gives: \(x = \frac{-0 \pm \sqrt{0^2-4*(5)*35}}{2*5}\), which simplifies to \(x = \pm \sqrt{-140}\). However, we can stop here as negative numbers can't be square rooted in real numbers, indicating there's no real root for the quadratic equation in this case.

Solve using another method (Factoring)

Let's solve the equation \(5x^2+35=0\) by factoring. First, remove common factor which is 5: \(5(x^2 + 7)=0\). Equating each factor to zero and solving for x, we get \(5=0\) (which isn't possible) and \(x^2+7=0\). Solving for x in \(x^2+7=0\), we find there's also no real root for this equation. This matches the result obtained using the Quadratic Formula.

Comparison and Preference

The Quadratic Formula is more advantageous when the equation is complex or when it cannot be easily factored. Factoring is often easier to use when the quadratic equation can be easily broken down into factors. In this case, both methods have correctly determined that there are no real roots. Considering this, the preference depends on the individual. If the student finds it easier to use the Quadratic Formula, they may choose to use it. If they find factoring simpler, they can use that method.

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. It works for any quadratic equation of the form: \[ ax^2 + bx + c = 0 \]To use the quadratic formula, you can plug the values of the coefficients \(a\), \(b\), and \(c\) into the following formula:\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]This formula calculates the roots of the quadratic equation, which are the values of \(x\) that satisfy the equation.

• "\(-b\)" acts to move or shift the solutions horizontally on the x-axis.
• "\(\pm \sqrt{b^2-4ac}\)" introduces the concept of two potential solutions due to the plus or minus sign.
• "\(2a\)" normalizes the expression by dividing the adjustments made by \(a\).

The discriminant, "\(b^2 - 4ac\)", tells us about the nature of the roots.
If it is positive, the roots are real; if it is zero, there is exactly one real root. If it is negative, the roots are complex, as seen in the problem above.

Factoring

Factoring is another technique to solve quadratic equations, especially when they can be easily broken down into simpler components. The goal is to express the quadratic in the form of a product of two binomials. For example, if you have an equation \(x^2 + 7 = 0\), you start by factoring out any common terms.
In the given example, the common factor was 5, so the equation simplified to \(5(x^2 + 7) = 0\).

• Firstly, factor out the greatest common factor (GCF) from the equation components.
• Next, you equate the expression to zero, and find the solutions by setting each factor to zero.

In some cases, like \(x^2 + 7 = 0\), factoring still reveals that there are no real roots. When the expression is unfactorable within integers or rational numbers, it indicates potential complex solutions.

Complex Numbers

Complex numbers come into play when equations have no real roots, which happens when the discriminant is negative. In such instances, solutions are not on the real number line but involve the imaginary unit "\(i\)".

• The imaginary unit "\(i\)" is defined as \(\sqrt{-1}\).
• A complex number has both a real part and an imaginary part and is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers.
• For the given problem, \(b^2-4ac = -140\), a negative value, meaning the solutions exist along the complex plane.

In the quadratic formula, encountering a negative under the square root (\(\sqrt{-140}\)) suggests complex roots, often written as \(\pm i\sqrt{140}\). It broadens the concept of numbers beyond the real values.

Real Roots

Real roots of a quadratic equation are the solutions that make the equation true for real numbers. Based on the discriminant (\(b^2-4ac\)), you can predict the existence of these roots:

• If the discriminant is positive, the quadratic will have two distinct real roots, meaning the parabola intersects the x-axis at two points.
• If it is zero, there is a single real root, and the parabola touches the x-axis at one point, known as a double root.
• However, if it is negative, as in the solved exercise, there are no real roots; it indicates the parabola does not intersect the x-axis.

Here, since \(b^2-4ac = -140\), the roots are not real. Understanding these principles helps predict and verify the nature of solutions based on the equation's components.