Question

What are the solution(s) to the following equation? \(5 x^{2}-15 x+10=0\) (A) 0 (B) 1,2 (C) 1,4 (D) 2,5

Short answer

Option (B) 1, 2.

Step-by-step solution

Identify Equation Type

The given equation is a quadratic equation in the standard form, which is written as \(ax^2 + bx + c = 0\). In this equation, we have: \(a = 5\), \(b = -15\), and \(c = 10\). To find the solutions, we can use the quadratic formula.

Apply the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). By substituting the values of \(a\), \(b\), and \(c\) into this formula, we get: \(x = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \times 5 \times 10}}{2 \times 5}\).

Simplify Inside the Square Root

Calculate \(b^2 - 4ac\): \((-15)^2 = 225\) and \(4 \times 5 \times 10 = 200\). So, \(b^2 - 4ac = 225 - 200 = 25\).

Solve the Square Root and Simplify

Calculate the square root: \(\sqrt{25} = 5\). Therefore, the quadratic formula becomes \(x = \frac{15 \pm 5}{10}\).

Compute the Two Solutions

Calculate the two possible solutions: \(x = \frac{15 + 5}{10} = 2\) and \(x = \frac{15 - 5}{10} = 1\). Thus, the solutions are \(x = 1\) and \(x = 2\).

Select the Correct Multiple Choice Answer

The solutions \(x = 1\) and \(x = 2\) correspond to option (B). Therefore, (B) 1, 2 is the correct answer.

Key concepts

Quadratic Formula

When faced with a quadratic equation, often the most efficient way to find the solutions is to use the quadratic formula. This formula offers a surefire method to solve any quadratic equation of the form \(ax^2 + bx + c = 0\). It is expressed as \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here’s how it works:

• The term \(-b\) involves changing the sign of the linear coefficient \(b\).
• \(\sqrt{b^2 - 4ac}\) represents the square root of the discriminant, a crucial part in determining the nature of the roots.
• The denominator \(2a\) normalizes the expression, factoring in the quadratic coefficient \(a\).

Using this formula, you can plug in the specific values of \(a\), \(b\), and \(c\) from your quadratic equation, like we did in solving \(5x^2-15x+10=0\). The resulting solutions can either be real, repeated, or complex depending on the discriminant.

Standard Form of a Quadratic Equation

Quadratic equations are typically presented in the standard form, which is \(ax^2 + bx + c = 0\). This form makes it easier to apply methods like the quadratic formula or factorization to find solutions. Let’s break down its components:

• \(a\) is the coefficient of \(x^2\). It cannot be zero because that would make the equation linear, not quadratic.
• \(b\) is the coefficient of \(x\). It can influence the axis of symmetry for the parabola when graphed.
• \(c\) is the constant term. It shifts the graph up or down on the coordinate plane.

Understanding this form is crucial for manipulating and solving quadratic equations. Once identified, you can easily substitute these coefficients into the quadratic formula and proceed with solving. In the equation \(5x^2-15x+10=0\), identifying \(a = 5\), \(b = -15\), and \(c = 10\) was the first step in finding the solutions.

Solving Equations

Solving quadratic equations involves determining the values of \(x\) that satisfy the equation. The quadratic formula is one method, but there are others worth mentioning:

• Factoring: If a quadratic can be expressed as a product of two binomials, solutions can often be found quickly by setting each binomial equal to zero.
• Completing the Square: This technique involves rearranging the equation to make one side a perfect square trinomial, which can then be solved by taking the square root.
• Graphing: The solutions to a quadratic are the \(x\)-coordinates where the graph of the equation crosses the \(x\)-axis.

Each method has its place, depending on the specific equation and context. For the equation \(5x^2-15x+10=0\), using the quadratic formula offered a straightforward path to the solutions \(x = 1\) and \(x = 2\). Familiarity with different solving methods enhances flexibility and efficiency when tackling quadratics in various mathematical problems.