Problem 326
Question
Solve. \(y^{2}-8 y+15=0\)
Step-by-Step Solution
Verified Answer
The solutions are y = 3 and y = 5.
1Step 1: Identify the quadratic equation
The given equation is a quadratic equation in the form of \[y^2 - 8y + 15 = 0\]
2Step 2: Factorize the quadratic equation
To factorize the quadratic equation, find two numbers that multiply to 15 (the constant term) and add up to -8 (the coefficient of the linear term). The numbers -3 and -5 satisfy these conditions.
3Step 3: Write the factored form
Rewrite the equation in its factored form using the numbers found in Step 2: \[(y - 3)(y - 5) = 0\]
4Step 4: Solve for y
Set each factor equal to zero and solve for y: \[y - 3 = 0\] which gives \[y = 3\] \[y - 5 = 0\] which gives \[y = 5\]
5Step 5: Write the solutions
The solutions to the quadratic equation are \[y = 3\] and \[y = 5\]
Key Concepts
FactorizationQuadratic FormulaSolving Quadratic Equations
Factorization
Factorization is an essential method for solving quadratic equations. It involves rewriting the equation as a product of simpler expressions. In the exercise, the quadratic equation given is \(y^{2}-8y+15=0\).
First, identify two numbers that multiply to the constant term 15 and simultaneously add up to the coefficient of the linear term -8. Here, the numbers -3 and -5 meet these criteria:
Using these numbers, the quadratic equation \(y^{2}-8y+15\) can be rewritten as two binomials: \( (y-3)(y-5)=0 \). This step decomposes the quadratic into simpler parts, making it easier to solve.
First, identify two numbers that multiply to the constant term 15 and simultaneously add up to the coefficient of the linear term -8. Here, the numbers -3 and -5 meet these criteria:
- -3 * -5 = 15
- -3 + -5 = -8
Using these numbers, the quadratic equation \(y^{2}-8y+15\) can be rewritten as two binomials: \( (y-3)(y-5)=0 \). This step decomposes the quadratic into simpler parts, making it easier to solve.
Quadratic Formula
The quadratic formula provides a standard way to solve any quadratic equation. If the equation is in the form \(ax^2 + bx + c = 0\), the solutions can be found using the formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
For our specific equation \(y^{2}-8y+15=0\), we can identify the coefficients:
Plugging these values into the quadratic formula, we get:
The solutions we obtain are the same as from factorizing: y = 3 and y = 5.
For our specific equation \(y^{2}-8y+15=0\), we can identify the coefficients:
- a = 1
- b = -8
- c = 15
Plugging these values into the quadratic formula, we get:
- y = \( \frac{8 \pm \sqrt{(-8)^2 - 4*1*15}}{2*1} \)
- y = \( \frac{8 \pm \sqrt{64 - 60}}{2} \)
- y = \( \frac{8 \pm \sqrt{4}}{2} \)
- y = \( \frac{8 \pm 2}{2} \)
- y = 5 \ | \-3
The solutions we obtain are the same as from factorizing: y = 3 and y = 5.
Solving Quadratic Equations
Solving quadratic equations involves finding the values of the variable that satisfy the equation. In this exercise, we used factorization:
This method is efficient and often simpler when the quadratic equation is factorable. For non-factorable equations, the quadratic formula is an excellent alternative. Both methods allow us to find the roots of the quadratic equation effectively.
- First, the equation \(y^{2}-8y+15\) was transformed into its factored form \( (y-3)(y-5)=0 \).
- Next, we set each factor equal to zero: \( y-3=0 \) and \( y-5=0 \).
- Solving these simple linear equations provided the solutions: \( y=3 \) and \( y=5 \).
This method is efficient and often simpler when the quadratic equation is factorable. For non-factorable equations, the quadratic formula is an excellent alternative. Both methods allow us to find the roots of the quadratic equation effectively.