Problem 68 - Algebra A Combined Function

Question

Solve each equation. $$ 4 x^{2}-20 x=-5 x^{2}-6 x-5 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{5}{9}\) and \(x = 1\).

Step 1: Arrange All Terms on One Side

Start by moving all terms to one side of the equation to form a single quadratic equation. Add \(5x^2 + 6x + 5\) to both sides of the equation to get: \[ 4x^2 - 20x + 5x^2 + 6x + 5 = 0 \] This simplifies to: \[ 9x^2 - 14x + 5 = 0 \]

Step 2: Factor the Quadratic Equation

Try to factor the quadratic equation \(9x^2 - 14x + 5 = 0\). Look for two numbers that multiply to \(9 \times 5 = 45\) and add up to \(-14\). These numbers are \(-9\) and \(-5\). The factorization is then: \[ (9x - 5)(x - 1) = 0 \]

Step 3: Solve for Each Variable

Set each factor equal to zero and solve for \(x\). For \(9x - 5 = 0\), add 5 to both sides and then divide by 9: \(x = \frac{5}{9}\). For \(x - 1 = 0\), add 1 to both sides: \(x = 1\).

Step 4: Verify the Solutions

Verify the solutions by substituting \(x = \frac{5}{9}\) and \(x = 1\) back into the original equation to ensure they satisfy it. Upon substitution, both values satisfy the quadratic equation, confirming the solutions.

Key Concepts

Factoring Quadratics

Factoring quadratics is a crucial skill in solving quadratic equations. It involves rewriting the quadratic expression as a product of two binomials. In our problem, we have the quadratic equation \[ 9x^2 - 14x + 5 = 0 \] To factor this, we need to find two numbers that multiply to the product of the coefficient of \(x^2\) term (9) and the constant term (5), which equals 45, and also add up to the coefficient of the \(x\) term (-14). - **Identify Products and Sums:** Here, we found that - the numbers -9 and -5 have a product of 45 - and they add up to -14 Once these numbers are identified, they help rewrite the middle term. The quadratic becomes:\[(9x - 5)(x - 1) = 0\]The equation is now factored into two simple linear factors that can be solved separately.

Solving Equations

Once we have factored the quadratic equation into \[(9x - 5)(x - 1) = 0\], solving these equations is straightforward. Each factor represents a potential solution for the equation. You just need to set each factor equal to zero and solve for \(x\). - **Solve Each Factor Separately:** 1. For the factor \((9x - 5)\), set it to zero: - \(9x - 5 = 0 \) - Add 5 to both sides: \(9x = 5\) - Divide by 9: \(x = \frac{5}{9}\) 2. For the factor \((x - 1)\), set it to zero: - \(x - 1 = 0 \) - Add 1 to both sides: \(x = 1\)These values, \(x = \frac{5}{9}\) and \(x = 1\), are potential solutions to the original quadratic equation.

Verification of Solutions

Verification is an important step to ensure that the solutions derived actually satisfy the original equation. For our equation \(4 x^2 - 20 x = -5 x^2 - 6 x - 5 \), we have found two potential solutions: \(x = \frac{5}{9}\) and \(x = 1\). To verify:- **Substitute Back:** Insert each value back into the original equation to check. 1. For \(x = \frac{5}{9}\): - Substitute in the equation and simplify both sides. If both sides equate, the solution is correct. 2. For \(x = 1\): - Similarly, substitute and check both sides.If both substitutions accurately balance the equation, it confirms that both \(x = \frac{5}{9}\) and \(x = 1\) are indeed valid solutions. This step helps in ensuring no errors were made during solving.