Problem 38 - Elementary Algebra

Question

(a) solve. (b) check. $$ y^{2}+9 y+18=0 $$

Step-by-Step Solution

Answer

The solutions are y = -3 and y = -6.

Step 1: Identify the quadratic equation components

The given quadratic equation is y^2 + 9y + 18 = 0Identify the coefficients from the equation: a = 1 (coefficient of y^2), b = 9 (coefficient of y), c = 18 (constant term).

Step 2: Apply the quadratic formula

To find the solutions for y, use the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plug in the values of a, b, and c: \[y = \frac{-9 \pm \sqrt{9^2 - 4 \times 1 \times 18}}{2 \times 1}\] \[y = \frac{-9 \pm \sqrt{81 - 72}}{2}\] \[y = \frac{-9 \pm \sqrt{9}}{2}\] \[y = \frac{-9 \pm 3}{2}\]

Step 3: Simplify to find the roots

Solve the two possible values for y: \[y = \frac{-9 + 3}{2} = \frac{-6}{2} = -3\] \[y = \frac{-9 - 3}{2} = \frac{-12}{2} = -6\]So the solutions are y = -3 and y = -6.

Step 4: Check the solutions

Substitute y = -3 into the original equation: \[(-3)^2 + 9(-3) + 18 = 0\] \[9 - 27 + 18 = 0\] \[0 = 0\]Substitute y = -6 into the original equation: \[(-6)^2 + 9(-6) + 18 = 0\] \[36 - 54 + 18 = 0\] \[0 = 0\]Both solutions satisfy the original equation.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is any equation that can be written in the standard form: a y^2 + b y + c = 0Here, 'a', 'b', and 'c' are constants, with 'a' not equal to zero. To solve for 'y', we can use the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives us the potential roots of the equation. First, identify the coefficients 'a', 'b', and 'c' in your quadratic equation. Insert these coefficients into the formula, and then solve for 'y'.

Roots of a Quadratic Equation

The solutions for 'y' that you find using the quadratic formula are known as the roots of the quadratic equation. These roots can be real or complex numbers. In the given exercise: y^2 + 9y + 18 = 0 the coefficients are a = 1, b = 9, and c = 18. By plugging these values into the quadratic formula, we calculated the following: \[y = \frac{-9 \pm \sqrt{81 - 72}}{2} = \frac{-9 \pm \sqrt{9}}{2} = \frac{-9 \pm 3}{2}\] This simplifies to two roots: \[ y = \frac{-9 + 3}{2} = -3\] \[ y = \frac{-9 - 3}{2} = -6\] Thus the roots of the quadratic equation y^2 + 9y + 18 = 0 are -3 and -6.

Checking Solutions in Algebra

After finding the roots of a quadratic equation, it's important to verify that they are correct. This step ensures the solutions satisfy the original equation. For y = -3 and y = -6, we substitute each root into the original equation: Substituting y = -3:\[(-3)^2 + 9(-3) + 18 = 0\] the equation simplifies as follows: \[9 - 27 + 18 = 0\] \[0 = 0\] Substituting y = -6:\[(-6)^2 + 9(-6) + 18 = 0\] the equation simplifies as follows: \[36 - 54 + 18 = 0\] \[0 = 0\] Since both solutions make the original equation true, we have successfully verified that y = -3 and y = -6 are indeed the roots of the equation y^2 + 9y + 18 = 0.