Problem 70 - Algebra for College Students

Question

Use the quadratic formula and a calculator to solve each equation. Round answers to three decimal places and check your answers. $$1.44 x^{2}+5.52 x+5.29=0$$

Step-by-Step Solution

Answer

The solutions are approximately \( x_1 = -1.869 \) and \( x_2 = -1.965 \).

Step 1 - Identify coefficients

Identify the coefficients from the quadratic equation \( ax^2 + bx + c = 0 \). Here, \( a = 1.44 \), \( b = 5.52 \), and \( c = 5.29 \).

Step 2 - Write the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Step 3 - Calculate the discriminant

Calculate the discriminant using \( b^2 - 4ac \).\[ b^2 - 4ac = 5.52^2 - 4 \cdot 1.44 \cdot 5.29 \]\[ = 30.4704 - 30.4512 \]\[ = 0.0192 \]

Step 4 - Calculate the solutions

Plug the coefficients and the discriminant into the quadratic formula and solve for \( x \).\[ x = \frac{-5.52 \pm \sqrt{0.0192}}{2 \cdot 1.44} \]First solution:\[ x_1 = \frac{-5.52 + \sqrt{0.0192}}{2 \cdot 1.44} \]\[ x_1 = \frac{-5.52 + 0.1386}{2.88} \]\[ x_1 = \frac{-5.3814}{2.88} \]\[ x_1 \approx -1.869 \]Second solution:\[ x_2 = \frac{-5.52 - \sqrt{0.0192}}{2 \cdot 1.44} \]\[ x_2 = \frac{-5.52 - 0.1386}{2.88} \]\[ x_2 = \frac{-5.6586}{2.88} \]\[ x_2 \approx -1.965 \]

Step 5 - Verify the solutions

Substitute \( x_1 \approx -1.869 \) and \( x_2 \approx -1.965 \) back into the original equation to check the solutions.For \( x_1 = -1.869 \):\[ 1.44(-1.869)^2 + 5.52(-1.869) + 5.29 \approx 0 \]For \( x_2 = -1.965 \):\[ 1.44(-1.965)^2 + 5.52(-1.965) + 5.29 \approx 0 \]

Key Concepts

solving quadratic equations

In algebra, solving quadratic equations is a vital skill. A quadratic equation is any equation in the form of \( ax^2 + bx + c = 0 \). Here, \( a, b, \) and \( c \) are constants, and \( x \) represents an unknown variable we aim to solve for.
One of the most reliable methods to solve quadratic equations is using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The '±' sign means there are two possible solutions.
Let's break down the process:

• Identify the coefficients \( a, b, \) and \( c \).
• Substitute these values into the quadratic formula.
• Calculate the discriminant, \( b^2 - 4ac \), to determine the solutions.
• Solve for \( x \) using the formula's two possible values.

discriminant calculation

The discriminant is an important part of the quadratic formula. It helps determine the nature and number of solutions of the quadratic equation.
The discriminant \( D \) is given by the expression inside the square root of the quadratic formula: \( D = b^2 - 4ac \).

Here’s how to interpret the discriminant:

• If \( D > 0 \), the equation has two distinct real solutions.
• If \( D = 0 \), the equation has exactly one real solution (a repeated root).
• If \( D < 0 \), the equation has two complex solutions (no real solutions).

In our example, the discriminant calculation was:
\[ D = 5.52^2 - 4 \cdot 1.44 \cdot 5.29 = 0.0192 \]
Since \( D > 0 \), we have two distinct real solutions.

verifying solutions

Verifying the solutions of a quadratic equation is the final and crucial step in the problem-solving process. This ensures the computed solutions are correct.
After finding the values of \( x \), substitute these back into the original quadratic equation to check if both sides of the equation balance.
Let's verify the solutions \( x_1 = -1.869 \) and \( x_2 = -1.965 \) found earlier:

• For \( x_1 = -1.869 \), substitute it back:
  \[ 1.44(-1.869)^2 + 5.52(-1.869) + 5.29 \approx 0 \]
  The left-hand side should be extremely close to zero if it's accurate.
• For \( x_2 = -1.965 \), substitute it back:
  \[ 1.44(-1.965)^2 + 5.52(-1.965) + 5.29 \approx 0 \]
  Again, the left-hand side should approach zero if the solution is correct.

Ensuring both solutions satisfy the original equation confirms their validity.