Problem 13
Question
Solve the equation by completing the square. \(x^{2}+2 x-5=0\)
Step-by-Step Solution
Verified Answer
The solutions are \( x = -1 \pm \sqrt{6} \).
1Step 1: Move the constant term to the other side
Start with the equation:\[ x^{2} + 2x - 5 = 0 \]Add 5 to both sides to isolate the zero on one side:\[ x^{2} + 2x = 5 \]
2Step 2: Find the term to complete the square
To complete the square, we need a perfect square trinomial on the left side. Take the coefficient of \(x\), divide it by 2, and square it:- Coefficient of \(x\) is 2.- \(\left(\frac{2}{2}\right)^{2} = 1\).So, you add and subtract 1 on the left side to complete the square:\[ x^{2} + 2x + 1 - 1 = 5 \]
3Step 3: Write the perfect square trinomial
Rewrite the expression as a perfect square trinomial by grouping:\[ (x^2 + 2x + 1) - 1 = 5 \]This can be rewritten as:\[ (x + 1)^{2} - 1 = 5 \]
4Step 4: Isolate the perfect square
Add 1 to both sides to isolate the perfect square:\[ (x + 1)^{2} = 6 \]
5Step 5: Solve for x using the square root property
Take the square root of both sides to solve for \(x\):\[ x + 1 = \pm \sqrt{6} \]
6Step 6: Solve for x
Subtract 1 from both sides to solve for \(x\):\[ x = -1 \pm \sqrt{6} \]
7Step 7: Conclusion
The solutions to the equation \(x^{2}+2x-5=0\) are \[ x = -1 + \sqrt{6} \] and \[ x = -1 - \sqrt{6} \].
Key Concepts
Perfect Square TrinomialQuadratic EquationSquare Root Property
Perfect Square Trinomial
A perfect square trinomial is a special type of quadratic expression. It takes the form of \((a+b)^2 = a^2 + 2ab + b^2\),where you can see it folds out into a square expression. This kind of trinomial is called 'perfect' because it can be rewritten as the square of a binomial.The process of converting a normal quadratic expression into a perfect square trinomial involves identifying the middle term (the term with 'x') and ensuring it matches the pattern for perfect square trinomials.
- Identify the linear coefficient (the number in front of 'x').
- Divide this coefficient by 2.
- Square the result.
Quadratic Equation
Quadratic equations are polynomials that have the form\(ax^2 + bx + c = 0\),with 'a', 'b', and 'c' as coefficients. Solving these equations can seem daunting, but remember they always represent a curve known as a parabola. Completing the square is one method used to solve these equations. When no apparent or easy factorization exists, transforming the quadratic into a perfect square trinomial can make the equation easier to handle, ultimately leading us to a solution. This evolves the equation into something reminiscent of the format \((x + p)^2 = q\).In our example, the original equation \(x^2 + 2x - 5 = 0\)was a typical quadratic equation. Through completing the square, we transformed it into \((x+1)^2 = 6\),much easier to solve, paving the way for applying the square root property.
Square Root Property
The square root property is a valuable tool for solving equations that involve a squared term set equal to a constant. Once the quadratic expression is transformed into a square of a binomial, we can use the square root function to further simplify and find solutions for 'x'.In simple terms, if you have an equation like \((X)^2 = k\),you can apply the square root to both sides:\(X = \pm \sqrt{k}\).Don't forget the plus-minus symbol; it accounts for two possible solutions.In our example, once we isolated the squared term \((x + 1)^2 = 6\),we applied the square root property:
- Square root both sides.
- This gives two possible equations: \(x + 1 = \sqrt{6}\) and \(x + 1 = - \sqrt{6}\).
Other exercises in this chapter
Problem 13
Evaluate the expression and write the result in the form \(a+b i .\) $$ (-6+6 i)+(9-i) $$
View solution Problem 13
1–54 ? Find all real solutions of the equation. $$ x^{3}-x^{2}+x-1=x^{2}+1 $$
View solution Problem 13
\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ x-3=2 x+6 $$
View solution Problem 14
Solve the equation. $$ 20+|2 x-4|=15 $$
View solution