Question

Find the real solutions of each equation. $$ 3(1-y)^{2}+5(1-y)+2=0 $$

Short answer

The real solutions are \(y = \frac{5}{3}\) and \(y = 2\).

Step-by-step solution

Introduce a substitution

Let’s use the substitution method. Define a new variable to simplify the equation. Let’s set \(u = 1 - y\). Substituting this in, the equation becomes \(3u^2 + 5u + 2 = 0\).

Apply the quadratic formula

The quadratic formula is given by \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 3\), \(b = 5\), and \(c = 2\). Substitute these values into the quadratic formula to get \(u = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}\).

Simplify under the square root

Simplify the expression under the square root: \(5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1\). So the equation now is \(u = \frac{-5 \pm 1}{6}\).

Find the values for \(u\)

Solve for \(u\):\(u = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3}\)and \(u = \frac{-5 - 1}{6} = \frac{-6}{6} = -1\).

Substitute back to find \(y\)

Recall that \(u = 1 - y\). Now substitute back to find the values of \(y\).For \(u = -\frac{2}{3}\): \( -\frac{2}{3} = 1 - y \ y = 1 + \frac{2}{3} = \frac{5}{3}\).For \(u = -1\): \( -1 = 1 - y \ y = 1 + 1 = 2\).

Key concepts

substitution method in algebra

The substitution method is a powerful algebraic tool. It involves replacing one part of an equation with a new variable. This makes the equation easier to solve. For example, if you face a complex equation like \(3(1-y)^2 + 5(1-y) + 2 = 0\), you can introduce a new variable. In our case, let \(u = 1 - y\). This substitution transforms the original equation into a simpler quadratic form: \(3u^2 + 5u + 2 = 0\). Notice how we changed a slightly complicated expression into something we can handle easily.
Always remember to revert back to the original variables after solving. So, once you find the values of \(u\), you need to re-substitute them back into \(u = 1 - y\) to find \(y\). That way, the solution still answers the original problem.

quadratic formula

The quadratic formula is essential for solving quadratic equations. It is given by:
\ \u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\ \
where \(a, b,\) and \(c\) are coefficients from the quadratic equation of the form \(ax^2 + bx + c = 0\).
To use the formula, you need to plug in the values of \(a\), \(b\), and \(c\) from your equation. For example, in \(3u^2 + 5u + 2 = 0\), we have \(a = 3\), \(b = 5\), and \(c = 2 \). Substituting these into the formula yields:
\ \u = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}\ \
Simplifying it results in two possible values for \(u\). These values represent the solutions to our simplified quadratic equation. Remember, the discriminant (the part under the square root, \(b^2 - 4ac\)) determines the nature of the roots. If it's positive, you get two distinct real solutions.

solving quadratic equations

Solving quadratic equations involves multiple steps. Here's a simplified breakdown:

• First, bring the equation to the standard quadratic form \(ax^2 + bx + c = 0\).
  
• Next, identify the coefficients \(a\), \(b\), and \(c\).
• Then, apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Simplify the expression under the square root (discriminant).
• Calculate the two possible values of \(x\).
• Finally, interpret these values in the context of the original problem.

In our given problem, we introduced a substitution \(u = 1 - y\) to simplify solving. After finding \(u\)’s values with the quadratic formula, we substituted back to find \(y\). This method ensures that we correctly answer the original problem. Understanding these steps helps you tackle real-world problems involving quadratic equations efficiently.