SAT Quadratic Equations Mastery Quiz (Hard)

Expanding the equation, we get \( 2u^2 + 3u - 7 = 3u^2 - 12u + 10u - 21 \). Simplifying, we obtain \( 2u^2 + 3u - 7 = 3u^2 - 2u - 21 \). Rearranging terms, we find \( -u^2 + 5u + 14 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = -1 \) and \( b = 5 \), which yields \( -5 \).

Expanding the equation, we get \( 4t^2 + 4t - 6 = 3t^2 - 6t + 8t - 10 \). Simplifying, we obtain \( 4t^2 + 4t - 6 = 3t^2 + 2t - 10 \). Rearranging terms, we find \( t^2 + 2t + 4 = 0 \). The product of the solutions is given by \( c/a \), where \( a = 1 \) and \( c = 4 \), which yields \( 4 \).

Expanding the equation, we get \( 3z^2 - 4z - 8 = 2z^2 - 6z + 5z - 16 \). Simplifying, we obtain \( 3z^2 - 4z - 8 = 2z^2 - z - 16 \). Rearranging terms, we find \( z^2 + 3z + 8 = 0 \). Using the quadratic formula, the solutions are \( z = frac{-3 pm sqrt{9 - 32}}{2} \), yielding complex numbers. The difference between the real parts is \( 0 \).

Expanding the equation, we get \( 2y^2 + 5y - 15 = 3y^2 - 9y + 2y - 10 \). Simplifying, we obtain \( 2y^2 + 5y - 15 = 3y^2 - 7y - 10 \). Rearranging terms, we find \( -y^2 + 12y - 5 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = -1 \) and \( b = 12 \), which yields \( 12 \).

Expanding the equation, we get \( 3x^2 + 6x = 2x^2 + 7x - 10 \). Rearranging terms, we obtain \( x^2 - x + 10 = 0 \). The product of the solutions is given by \( c/a \), where \( a = 1 \) and \( c = 10 \), which yields \( 10 \).

Expanding the equation, we get \( p^2 - 3p - 2 = p^2 - 8p + 10 \). Rearranging terms, we find \( 5p - 12 = 0 \). Since this equation does not form a quadratic, there are no solutions in the traditional sense; hence, the product of the solutions is undefined.

Expanding the equation, we get \( n^2 + 4n - 5 = 2n^2 - 4n + 7 \). Rearranging terms, we obtain \( -n^2 + 8n - 12 = 0 \). Using the quadratic formula, the solutions are \( n = frac{-8 pm sqrt{64 - 48}}{-2} \), yielding \( n = frac{8 pm sqrt{16}}{2} \). The difference between the solutions is \( 4 \).

Expanding the equation, we get \( 2m^2 + 3m = m^2 + 10m + 21 \). Rearranging terms, we obtain \( m^2 - 7m - 21 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = 1 \) and \( b = -7 \), which yields \( 7 \).

Expanding the equation, we get \( 3v^2 - 4v - 1 = 4v^2 - 12v + 5 \). Rearranging terms, we obtain \( -v^2 + 8v - 6 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = -1 \) and \( b = 8 \), which yields \( 8 \).

Expanding the equation, we get \( u^2 + 5u - 10 = 2u^2 + 15u - 20 \). Rearranging terms, we find \( -u^2 - 10u + 10 = 0 \). The product of the solutions is given by \( c/a \), where \( a = -1 \) and \( c = 10 \), which yields \( -10 \).

Expanding the equation, we get \( 4t^2 - 4t - 3 = t^2 + 7t - 10 \). Rearranging terms, we obtain \( 3t^2 - 11t + 7 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = 3 \) and \( b = -11 \), which yields \( 11/3 \).

Expanding the equation, we get \( 3w^2 - 2w = 5w^2 - 20w + 12 \). Rearranging terms, we obtain \( -2w^2 + 18w - 12 = 0 \). The sum of the solutions is given by \( -b/a \), where \( a = -2 \) and \( b = 18 \), which yields \( 9 \).

Expanding the equation, we obtain \( 2z^2 + 6z = z^2 - 12z + 16 \). Rearranging terms, we get \( z^2 + 18z - 16 = 0 \). Using the quadratic formula, the solutions are \( z = frac{-18 pm sqrt{324 + 64}}{2} \), yielding \( z = frac{-18 pm sqrt{388}}{2} \). The difference between the solutions is \( sqrt{388} \).

Expanding the equation, we get \( 2y^2 + 5y - 7 = 3y^2 - 6y \). Rearranging terms, we obtain \( -y^2 + 11y - 7 = 0 \). Using the quadratic formula, the solutions are \( y = frac{-11 pm sqrt{121 - 28}}{-2} \), giving us \( y = frac{11 pm sqrt{93}}{2} \).

Expanding the equation, we get \( 3x^2 - 12x = 5x^2 - 20x - 8 \). Rearranging terms, we find \( -2x^2 + 8x + 8 = 0 \). The product of the solutions is given by \( c/a \), where \( a = -2 \) and \( c = 8 \), which yields \( -4 \).