Question

Determine whether each value of \(x\) is a solution of the equation. Equation $$ (x+5)(x-3)=20 $$ Values (a) \(x=3\) (b) \(x=-2\) (c) \(x=0\) (d) \(x=-7\)

Short answer

Given values, only \(x=-7\) is a solution to the equation. Other values are not solutions to the equation.

Step-by-step solution

Understand the equation

The given equation is \((x+5)(x-3)=20\). To find if a given value of \(x\) is a solution, we need to substitute \(x\) in the equation with the given value and check if the left side equals the right side.

Verify the solution for \(x = 3\)

For \(x = 3\), substitute 3 in the given equation to get \((3+5)(3-3)\) which simplifies to \(8 \cdot 0 = 0\). This is not equal to 20, so \(x = 3\) is not a solution of the equation.

Verify the solution for \(x = -2\)

For \(x = -2\), substitute -2 in the given equation to get \((-2+5)(-2-3)\) which becomes \((3)(-5) = -15\). This is not equal to 20, so \(x = -2\) is not a solution of the equation.

Verify the solution for \(x = 0\)

For \(x = 0\), substitute 0 in the given equation to get \((0+5)(0-3)\) which simplifies to \(5 \cdot -3 = -15\). This is not equal to 20, so \(x = 0\) is not a solution of the equation.

Verify the solution for \(x = -7\)

For \(x = -7\), substitute -7 in the given equation to get \((-7+5)(-7-3)\) which simplifies to \(-2 \cdot -10 = 20\). This is equal to 20, so \(x = -7\) is a solution of the equation.

Key concepts

Solution Verification

Solution verification is an essential part of solving algebraic equations. Once an equation is presented, we often need to test potential solutions. For the given equation \((x+5)(x-3)=20\), potential solutions for \(x\) must satisfy the equation. By substituting these values into the equation, we determine if they make the equation true.With solution verification, we proceed with these steps:

• Substitute the tested value of \(x\) into the equation.
• Calculate the resulting expression on the left-hand side.
• Compare it with the right-hand side value.

If both sides are equal, the tested value is indeed a solution. If they are not, it is not a solution. This method is straightforward and vital to confirm or disprove an assumption about potential solutions.

Substitution Method

The substitution method is a widely used technique in algebra for solving equations. It involves replacing a variable with a given value to check if the equation holds true. Let's see how this works in practice with the equation \((x+5)(x-3)=20\).For each potential value:

• Replace \(x\) in the equation with the specific number.
• Perform arithmetic operations to simplify both sides of the equation.
• Check if the result equals the expected outcome.

For example, substituting \(x = 3\) we find that the equation becomes \((3+5)(3-3)\), which simplifies to \(8 \cdot 0 = 0\). Since \(0\) is not equal to \(20\), \(x = 3\) is not a solution. Through substitution, we effectively test each value to see if it satisfies the original equation.

Quadratic Equations

Quadratic equations are a type of polynomial equation that take the form \(ax^2 + bx + c = 0\). However, the equation \((x+5)(x-3) = 20\) can also be associated with quadratic equations. By expanding the left side, we obtain a standard quadratic form:\[(x+5)(x-3) = x^2 - 3x + 5x - 15 = x^2 + 2x - 15\]To transform our initial problem into solving a quadratic equation, we bring all terms to one side:\[x^2 + 2x - 15 = 20\]Subtract \(20\) from both sides to set it to zero:\[x^2 + 2x - 35 = 0\]By tackling it as a quadratic equation, you can alternatively look for solutions using methods such as factoring, completing the square, or the quadratic formula. Although not strictly necessary for solution verification, understanding these interpretations provides a deeper insight into the nature of the equation and its solutions.