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Magazine Chapter 4: Problem 29
Solve. \(\quad(6 x+1)(x+1)=6\)
Short Answer
Expert verified
The solutions are \(x = \frac{1}{2}\) and \(x = \frac{-5}{3}\).
Step by step solution
01
Expand the Equation
The given equation is \((6x+1)(x+1)=6\). First, expand the left-hand side. Use the distributive property to multiply each term in the first binomial by each term in the second binomial. \[\begin{align*}(6x+1)(x+1) &= 6x \cdot x + 6x \cdot 1 + 1 \cdot x + 1 \cdot 1 \&= 6x^2 + 6x + x + 1 \&= 6x^2 + 7x + 1\end{align*}\]
02
Set the Equation to Zero
Now we have \(6x^2 + 7x + 1 = 6\). To solve this quadratic equation, first move 6 to the left side of the equation to set it to zero.\[6x^2 + 7x + 1 - 6 = 0\]Simplify the equation:\[6x^2 + 7x - 5 = 0\]
03
Apply the Quadratic Formula
The quadratic formula is given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For our equation \(6x^2 + 7x - 5 = 0\), the coefficients are: \(a = 6\), \(b = 7\), and \(c = -5\). Substitute these values into the quadratic formula:\[x = \frac{-7 \pm \sqrt{7^2 - 4 \times 6 \times (-5)}}{2 \times 6}\]Calculate the discriminant:\[x = \frac{-7 \pm \sqrt{49 + 120}}{12}\]\[x = \frac{-7 \pm \sqrt{169}}{12}\]
04
Solve for x
Simplify the square root and solve:\(\sqrt{169} = 13\)This gives two solutions:\[x = \frac{-7 + 13}{12} = \frac{6}{12} = \frac{1}{2}\]\[x = \frac{-7 - 13}{12} = \frac{-20}{12} = \frac{-5}{3}\]
05
Verify the Solutions
Substitute the solutions back into the original equation to verify.For \(x = \frac{1}{2}\):\[LHS = (6(\frac{1}{2})+1)(\frac{1}{2}+1) = (3+1)(\frac{3}{2}) = 4 \times \frac{3}{2} = 6\]For \(x = \frac{-5}{3}\):\[LHS = (6(\frac{-5}{3})+1)((\frac{-5}{3})+1) = (-10+1)(\frac{-2}{3}) = -9 \times \frac{-2}{3} = 6\]Both solutions satisfy the original equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations that are not easily factored. Given a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), the quadratic formula provides a solution for \( x \). The formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use the formula effectively, identify the coefficients \( a \), \( b \), and \( c \) from the equation. These coefficients are essential inputs for the formula. Here鈥檚 a quick rundown:
- \( a \) is the coefficient of \( x^2 \)
- \( b \) is the coefficient of \( x \)
- \( c \) is the constant term
Distributive Property
The distributive property is a fundamental algebraic principle that allows you to multiply a sum by distributing the multiplication across each term within a parenthesis. This property is crucial when expanding expressions like \((a + b)(c + d)\). Imagine each term of the first binomial multiplying each term of the second binomial:\[(a + b)(c + d) = ac + ad + bc + bd\]In our example, the equation \((6x+1)(x+1) = 6\), apply the distributive property as follows:
- Multiply \(6x\) by \(x\) and \(1\)
- Similarly, multiply \(1\) by \(x\) and \(1\)
Discriminant
The discriminant is a key part of the quadratic formula, given by \( b^2 - 4ac \). It helps determine the nature and number of roots for a quadratic equation. Here's how:
- If \( b^2 - 4ac > 0 \), there are two real and distinct solutions.
- If \( b^2 - 4ac = 0 \), there is one real, repeated solution.
- If \( b^2 - 4ac < 0 \), there are no real solutions, but two complex ones.
Verify Solutions
After solving a quadratic equation, it鈥檚 important to verify the solutions by plugging them back into the original equation. This step ensures the solutions are correct. For example, with solutions \( x = \frac{1}{2} \) and \( x = \frac{-5}{3} \) from the equation \((6x+1)(x+1) = 6\):
- Substitute \( x = \frac{1}{2} \) into the equation. The left-hand side (LHS) simplifies to 6, matching the right-hand side (RHS).
- Substitute \( x = \frac{-5}{3} \) into the equation. Again, LHS simplifies to 6, confirming the solution.
