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Chapter 3: Problem 63

Solve each quadratic equation by completing the square. $$2 x^{2}+6 x-3=0$$

Short Answer

Expert verified
\(x = -\frac{3}{2} \pm \frac{\sqrt{15}}{2}\)

Step by step solution

01

Move Constant to the Other Side

Start with the quadratic equation \(2x^2 + 6x - 3 = 0\). Move the constant term \(-3\) to the other side of the equation to isolate the quadratic and linear terms: \[2x^2 + 6x = 3\]
02

Divide by Coefficient of x虏

To make the coefficient of \(x^2\) equal to 1, divide every term in the equation by 2:\[x^2 + 3x = \frac{3}{2}\]
03

Complete the Square

Find the number needed to complete the square, which is \(\left(\frac{b}{2}\right)^2\), where \(b\) is the coefficient of \(x\). Here, \(b = 3\), so:\[\left(\frac{3}{2}\right)^2 = \frac{9}{4}\]Add this value inside the equation and remember to balance the equation by adding the same value to the other side:\[x^2 + 3x + \frac{9}{4} = \frac{3}{2} + \frac{9}{4}\]
04

Simplify the Equation

Simplify the right side of the equation:\[\frac{3}{2} = \frac{6}{4}\], so:\[\frac{6}{4} + \frac{9}{4} = \frac{15}{4}\]The equation becomes:\[x^2 + 3x + \frac{9}{4} = \frac{15}{4}\]
05

Write as a Perfect Square

The left side of the equation now is a perfect square trinomial. Write it in factored form:\[(x + \frac{3}{2})^2 = \frac{15}{4}\]
06

Solve for x

Take the square root of both sides of the equation:\[x + \frac{3}{2} = \pm \sqrt{\frac{15}{4}}\]This simplifies to:\[x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2}\]Subtract \(\frac{3}{2}\) from both sides to isolate \(x\):\[x = -\frac{3}{2} \pm \frac{\sqrt{15}}{2}\]
07

Write the Final Answer

Express the two potential solutions for \(x\):\[x = -\frac{3}{2} + \frac{\sqrt{15}}{2}\]\[x = -\frac{3}{2} - \frac{\sqrt{15}}{2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Completing the Square
Completing the square is a method used to solve quadratic equations, which allows us to convert a standard quadratic equation into a form that reveals its roots more clearly. The process involves creating a perfect square trinomial from the original quadratic equation. This demonstration takes a quadratic equation of the form \(ax^2 + bx + c = 0\) and manipulates it to \((x + d)^2 = e\). Then, we can easily find the solutions for \(x\).

Here are the steps:
  • First, if the leading coefficient \(a\) of \(x^2\) is not 1, divide every term by \(a\) to make it 1.
  • Next, identify the coefficient of \(x\), which is \(b\), and compute \(\left(\frac{b}{2}\right)^2\).
  • Add and subtract this square within the equation to maintain balance.
  • Finally, rewrite the equation using the perfect square trinomial on one side.
Completing the square not only helps solve equations but also provides a foundation for understanding the derivation of the quadratic formula.
Solving Quadratic Equations
Solving quadratic equations can sometimes feel daunting, but with the right techniques, it becomes a manageable task. Quadratic equations are usually of the form \(ax^2 + bx + c = 0\), and solving them means finding the value(s) of \(x\) that satisfies the equation.

There are three popular methods for solving these equations:
  • Factoring the quadratic when possible.
  • Using the quadratic formula, which is derived from completing the square.
  • Completing the square to express the equation in vertex form.
These solutions are found considering the nature of the discriminant \(b^2 - 4ac\) contained in the quadratic formula:
  • If the discriminant is positive, there are two unique real solutions.
  • If the discriminant is zero, there is one real solution (a repeated root).
  • If the discriminant is negative, the solutions are complex or imaginary.
In our given exercise, completing the square was used because it elegantly converts the equation into a form where roots can be directly extracted.
Factoring Perfect Square Trinomials
Factoring perfect square trinomials is a key skill in algebra that simplifies the process of solving equations through completing the square. Essentially, a perfect square trinomial is the square of a binomial. When we have a trinomial that fits the pattern \(x^2 + 2xy + y^2\), it can be factored into \((x+y)^2\).

To recognize a perfect square trinomial, check these conditions:
  • The first term and last term are squares, i.e., \(a^2\) and \(b^2\).
  • The middle term is twice the product of the square roots of the first and last terms, i.e., \(2ab\).
In the original equation \[2x^2 + 6x - 3 = 0\],through completing the square, it was transformed into\[(x + \frac{3}{2})^2 = \frac{15}{4}\],allowing it to be easily solved. Understanding how to factor these trinomials makes solving quadratics faster and is crucial for simplifying expressions in algebra.

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Most popular questions from this chapter

Use the concepts of this section. Explain why a polynomial function of degree 4 with real coefficients has either zero, two, or four real zeros (counting multiplicities).

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=6 x^{3}+25 x^{2}+3 x-4 ; \quad k=-4$$

Use Descartes' rule of signs to determine the possible numbers of positive and negative real zeros for \(P(x) .\) Then, use a graph to determine the actual numbers of positive and negative real zeros. $$P(x)=x^{3}+2 x^{2}+x-10$$

Use the boundedness theorem to show that the real zeros of \(P(x)\) satisfy the given conditions. \(P(x)=x^{5}+2 x^{3}-2 x^{2}+5 x+5\); no real zero less than \(-1\)

Factor \(P(x)\) into linear factors given that \(k\) is a zero of \(P\). $$P(x)=2 x^{3}-3 x^{2}-17 x+30 ; \quad k=2$$
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