Question

Challenge Problems. $$7 r-2 r(2 r-3)-2=2 r^{2}-(r-2)(3+6 r)-8$$

Short answer

The equation has no real solutions as the discriminant is negative.

Step-by-step solution

Expand and Simplify Both Sides

Expand and simplify both sides of the equation. On the left side, distribute \( -2r \) to \( (2r-3) \) which gives us \( 7r - 4r^2 + 6r \). Combine like terms to get \(-4r^2 + 13r\). On the right side, expand \( -(r-2)(3+6r) \) to get \( -3r - 6r^2 + 6 + 12r \) and combine like terms to get \(-6r^2 + 9r + 6\). So we have the simplified equation \( -4r^2 + 13r - 2 = -6r^2 + 9r - 8 \).

Move All Terms to One Side

Move all terms to one side to set the equation equal to 0. Add \( 4r^2 \) to both sides and subtract \( 9r \) and add \( 2 \) from both sides to get \(0 = -2r^2 - 4r - 6\).

Simplify the Equation

Divide every term by \( -2 \) to simplify the equation, resulting in \(0 = r^{2}+2r+3\).

Solve the Quadratic Equation

Since the equation is now in the form \( r^{2}+2r+3=0 \), we can try to factorize it, complete the square, or use the quadratic formula. However, this equation does not have real solutions, as the discriminant \( b^2-4ac = 2^2 - 4(1)(3) = 4 - 12 = -8 \) is negative. So, there are no real solutions to the equation.

Key concepts

Expand and Simplify Equation

Expanding and simplifying equations serve as the foundation for solving complex algebraic expressions, particularly quadratic equations. To 'expand' typically means to multiply out brackets. For instance, if you have a form such as \(a(b+c)\), you would apply the distributive property to obtain \(ab + ac\). Simplifying involves combining like terms, which are terms with the same variable raised to the same power.

In our exercise, the initial step requires expanding both sides of the equation separately. On the left, \( -2r \times (2r-3) \) expands to \( -4r^2 + 6r \) and when combined with the existing \( 7r \) gives us \( -4r^2 + 13r \). On the right, \( -(r-2)(3+6r) \) expands to \( -3r - 6r^2 + 6 + 12r \) and simplifies to \( -6r^2 + 9r + 6 \). These steps consolidate the equation, making it easier to identify like terms and set the stage for the next phase in solving the quadratic equation.

Move Terms to One Side

To solve a quadratic equation, it is essential to first arrange the equation into a standard form. The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.

In our current example, after expanding and simplifying, we need to move all terms to one side to obtain that form. We do this by performing equivalent operations on both sides, such as adding \( 4r^2 \) and subtracting \( 9r \) and adding \( 2 \) to each side of the equation. The aim is to get a zero on one side, which is the basis for applying further solutions methods like factoring or using the quadratic formula. After arranging the terms, we end up with the equation \( 0 = -2r^2 - 4r - 6 \). Simplifying this by dividing by \( -2 \) is beneficial because it further reduces the equation to its simplest form, which, in this case, is \( r^2 + 2r + 3 = 0 \)—ready for the final step of actually solving the equation for \( r \).

Quadratic Formula

When factoring is difficult or impossible, the quadratic formula provides a reliable method for finding the roots of a quadratic equation. The formula is \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) where \( a \) is the coefficient of \( x^2 \) term, \( b \) is the coefficient of \( x \) term, and \( c \) is the constant term. Crucially, the discriminant, \( b^2-4ac \) inside the square root determines the nature of the roots: if positive, there are two real solutions; if zero, one real solution; if negative, as in the exercise, there are no real solutions.

Applying the quadratic formula to the simplified equation \( r^2 + 2r + 3 = 0 \) from our previous step, we compute the discriminant as \( 2^2 - 4 \times 1 \times 3 \) which gives us \( -8 \). Since the discriminant is negative, it indicates that this particular quadratic equation does not have real solutions. This finding speaks to the power of the quadratic formula as a comprehensive tool for solving quadratic equations, capable of indicating not just the solutions but their nature as well.