Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 9 x^{2}+8 x=5 $$

Short answer

The real solutions are \( x = \frac{-4 + \sqrt{61}}{9} \) and \( x = \frac{-4 - \sqrt{61}}{9} \).

Step-by-step solution

Write Down the Quadratic Formula

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

Identify Coefficients

From the equation \( 9x^2 + 8x = 5 \), identify the coefficients as follows: \( a = 9 \), \( b = 8 \), and \( c = -5 \) (Note that the equation needs to be written in standard form \( ax^2 + bx + c = 0 \) by subtracting 5 from both sides).

Plug the Coefficients into the Formula

Substitute \( a = 9 \), \( b = 8 \), and \( c = -5 \) into the quadratic formula:\[ x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 9 \cdot (-5)}}{2 \cdot 9}\]

Simplify Inside the Square Root

Calculate the discriminant \( b^2 - 4ac \):\[ 8^2 - 4 \cdot 9 \cdot (-5) = 64 + 180 = 244\]

Simplify the Entire Expression

Continue simplifying the expression:\[ x = \frac{-8 \pm \sqrt{244}}{18}\]

Further Simplify the Square Root

Simplify \( \sqrt{244} \):\[ \sqrt{244} = \sqrt{4 \cdot 61} = 2\sqrt{61}\]Now substitute back in:\[ x = \frac{-8 \pm 2\sqrt{61}}{18}\]

Reduce the Fraction

Divide the numerator and the denominator by 2:\[ x = \frac{-4 \pm \sqrt{61}}{9}\]

Key concepts

quadratic equation

Quadratic equations are a core part of algebra. They have the form \[ ax^2 + bx + c = 0 \] where 'a', 'b', and 'c' are constants, and 'x' represents the variable. The term 'quadratic' refers to the highest exponent of the variable, which is 2. When solving quadratic equations, we often use the quadratic formula. This helps us find the values of 'x' that satisfy the equation. The most commonly known quadratic equation solution method is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] The quadratic formula provides solutions (roots) that can be either real or complex numbers.

discriminant

The discriminant is a key part of the quadratic formula, located inside the square root (radical symbol). It is represented by the expression \[ D = b^2 - 4ac \]The value of the discriminant tells us the nature of the solutions of the quadratic equation:

• If the discriminant is greater than zero, there are two distinct real solutions.
• If the discriminant equals zero, there is exactly one real solution.
• If the discriminant is less than zero, there are no real solutions (the solutions are complex numbers).

In the original problem, the discriminant \[ b^2 - 4ac \] is calculated as: \[ 8^2 - 4 \times 9 \times (-5) = 64 + 180 = 244 \]Since 244 is greater than zero, this tells us that there are two distinct real solutions for the quadratic equation.

real solutions

Real solutions to a quadratic equation are the values of 'x' that make the equation true. Using the quadratic formula, we solve for 'x' by plugging in the appropriate values of 'a', 'b', and 'c'. From the exercise, substituting the coefficients \[ a = 9, \] \[ b = 8, \] and \[ c = -5 \] into the quadratic formula, we proceed with:
\[ x = \frac{-8 \pm \sqrt{244}}{18} \]
Simplifying further, we use \[ \sqrt{244} = 2\sqrt{61} \] to get:
\[ x = \frac{-8 \pm 2\sqrt{61}}{18} \]
Finally, by reducing the fraction, we find:
\[ x = \frac{-4 \pm \sqrt{61}}{9} \]
These are the real solutions to the quadratic equation \[ 9x^2 + 8x = 5 \] which means there are two distinct real solutions for 'x'.