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Chapter 0: Problem 73

Solve each quadratic equation by completing the square. $$x^{2}+4 x+1=0$$

Short Answer

Expert verified
The solutions for the quadratic equation are \( x = \sqrt{3}-2 \) and \( x = -\sqrt{3}-2 \).

Step by step solution

01

Arrange in general form

First, verify that the equation is in the general form of a quadratic equation, \( ax^2 + bx + c =0 \). In the given problem, \(a=1, b=4, c=1\). So the quadratic equation is \(x^2+4x+1=0\).
02

Make coefficient of x^2 equal to 1

Here the coefficient of \(x^2\) is 1, so we can proceed to the next step directly.
03

Complete the square

To complete the square, we take half the coefficient of x, square it, and add and subtract that number from the equation. In this case, half of the x-coefficient 4 is 2. Squaring 2 gives us 4. So, we will add and subtract 4 in our equation which results into \( (x^2+4x+4)-4 +1 = 0\). We can then write \( (x^2+4x+4) \) as \( (x+2)^2 \). Our transformed equation becomes \( (x+2)^2 -4 +1 = 0\).
04

Simplify and solve for x

Simplifying further, we get the equation \( (x+2)^2 -3 = 0\). Add 3 on both sides to isolate \( (x+2)^2 \) and get \( (x+2)^2 = 3\). Taking the square root of both parts leads to two possible solutions due to the \( 卤 \) sign: \( x+2 = \sqrt{3} \) or \( x+2 = - \sqrt{3} \). Subtracting 2 from both sides in each case, the solutions for the equation are \( x = \sqrt{3} -2 \) or \( x = - \sqrt{3} -2 \).

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Find \(b\) such that \(\frac{7 x+4}{b}+13=x\) will have a solution set given by \(\\{-6\\}\).

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