Question

Verify that $x=2$ and $x=3$ are both solutions of $x^2-5x+6=0$

Short answer

Question: Verify that both x=2 and x=3 are solutions of the equation $x^2-5x+6=0$. Answer: By substituting x=2 and x=3 into the equation $x^2-5x+6=0$ and simplifying, we found that the equation holds true for both x=2 and x=3. Thus, both x=2 and x=3 are solutions of the equation.

Step-by-step solution

Substitute $x=2$ into the equation.

First, let's substitute $x=2$ into the equation $x^2-5x+6=0$ and simplify to see if it is true: $(2{)}^2-5(2)+6=0$

Simplify with $x=2$

Now, let's simplify the equation when $x=2$: $4-10+6=0$

Verify the equation for $x=2$

By further simplification, we get: $0=0$ Since the left side of the equation equals the right side, we can conclude that $x=2$ is a solution of $x^2-5x+6=0$.

Substitute $x=3$ into the equation.

Similarly, let's substitute $x=3$ into the equation $x^2-5x+6=0$ and simplify to see if it is true: $(3{)}^2-5(3)+6=0$

Simplify with $x=3$

Now, let's simplify the equation when $x=3$: $9-15+6=0$

Verify the equation for $x=3$

By further simplification, we get: $0=0$ Since the left side of the equation equals the right side, we can conclude that $x=3$ is also a solution of $x^2-5x+6=0$. Since both $x=2$ and $x=3$ satisfy the equation $x^2-5x+6=0$, we have verified that they are both solutions of the equation.

Key concepts

Polynomial Factors

Understanding polynomial factors is crucial in algebra, particularly when solving quadratic equations. A factor of a polynomial is a binomial to which the polynomial can be divided evenly, without leaving a remainder.

For example, consider the quadratic equation from the exercise, $x^2-5x+6=0$. This equation has a special form where the polynomial on the left can be factored into the product of two binomials. In this case, the factors are $(x-2)(x-3)$. These factors are incredibly useful because they reveal the roots of the quadratic equation—which are values of $x$ that make the equation true, meaning they result in a zero value for the entire equation.

When we factor the polynomial, we apply the zero product property, which tells us if a product of factors equals zero, at least one of the factors must be zero. Hence, setting each binomial factor to zero, $x-2=0$ and $x-3=0$, gives us the roots, which are $x=2$ and $x=3$, confirming the solutions given in the exercise.

Roots of Quadratic

The roots of a quadratic equation are the values that satisfy the equation—when substituted for the variable, they yield zero. In the context of the given exercise, $x=2$ and $x=3$ are called roots because they satisfy the equation $x^2-5x+6=0$.

Finding these roots is a common goal in solving quadratic equations. There are multiple methods to find them, including factoring, completing the square, and using the quadratic formula. For simple quadratics like the one in our exercise, factoring is often the quickest method. However, for more complex equations where factoring isn't obvious, the quadratic formula, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, becomes very useful.

Knowing the roots of a quadratic equation has practical applications, such as determining the x-intercepts of a parabola in coordinate geometry. It's relevant for understanding the behavior of functions and finding solutions to a wide range of mathematical problems and real-world situations.

Algebraic Solutions

Algebraic solutions involve finding the values that satisfy an equation. In the exercise, we were given a quadratic equation $x^2-5x+6=0$, and we verified that $x=2$ and $x=3$ are solutions by substituting them into the equation and simplifying to show the equation holds true as it results in zero.

These algebraic solutions are not just random guesses. They are founded on solid mathematical principles, such as factorization, properties of equality, and the ability to manipulate and simplify expressions. When we substitute a value into the equation and simplify, we're confirming that this value is an actual solution to the equation.

Mastering algebraic solutions opens the door to solving more complex equations and understanding advanced topics. It's not limited to quadratics but applies to other forms of equations like linear, cubic, and even higher-degree polynomials. Each type of equation might require different techniques, but the core idea remains the same: algebraic solutions must satisfy the original equation completely.