Advanced Quadratic Equations Quiz (Medium) - SAT Math Practice Questions

Given \( b = 5 \) and \( c = 6 \), we have \( p + q = -5 \) and \( pq = 6 \). For the new equation with roots \( p^2 \) and \( q^2 \), the sum is \( p^2 + q^2 = (p+q)^2 - 2pq = (-5)^2 - 2(6) = 25 - 12 = 13 \), and the product is \( p^2 q^2 = (pq)^2 = 6^2 = 36 \). Thus, \( d = -13 \) and \( e = 36 \), so \( d + e = -13 + 36 = 23 \).

From \( x^2 - 6x + k = 0 \), we have \( m + n = 6 \) and \( mn = k \). The new roots \( m + 1 \) and \( n + 1 \) give the product \( (m+1)(n+1) = mn + m + n + 1 = k + 6 + 1 = k + 7 \). Therefore, \( q = k + 7 \).

Solving \( x^2 - 5x + 6 = 0 \) gives roots \( x = 2, 3 \). Solving \( y^2 - 4y + 4 = 0 \) gives root \( y = 2 \). Therefore, \( z = xy = 2 times 2 = 4 \) or \( z = 3 times 2 = 6 \). The possible values of \( z \) are \( 4 \) and \( 6 \).

Given \( b = 2a \), the sum of the roots for \( ax^2 + bx + c = 0 \) is \( -b/a = -2 \) and the product is \( c/a \). For \( cx^2 + bx + a = 0 \), the sum of the roots is \( -b/c = -2a/c \) and the product is \( a/c \). Since the roots are reciprocals, \( c/a = a/c \) implies \( a^2 = c^2 \). Given \( a neq c \), we have \( a = -c \), so \( a + c = 0 \).

Let the solutions be \( r_1 \) and \( r_2 \). We know \( r_1 + r_2 = 7/3 \) and \( r_1 r_2 = 2/3 \). The sum of the squares is \( r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1 r_2 = (7/3)^2 - 2(2/3) = 49/9 - 4/3 = 49/9 - 12/9 = 37/9 \).

For the equation to have two equal real roots, the discriminant must be zero: \( 4^2 - 4c = 0 \). Solving for \( c \) yields \( c = 4 \).

Using the sum of the roots formula, \( -b/a = 3 \), we substitute \( a = 4 \) and \( b = k \) to get \( -k/4 = 3 \). Solving for \( k \) gives \( k = -12 \).

Factorizing the equation \( x^2 - 5x + 6 = 0 \) results in \( (x-2)(x-3) = 0 \). The smallest integer value of \( x \) is \( 2 \).

Using the relationships \( -b/a \) and \( c/a \), we set up the equation \( -b/a = 2(c/a) \). Simplifying, we find \( -b = 2c \) or \( b = -2c \).

Using the roots \( 2 \) and \( -3 \), the equation can be formed as \( (x - 2)(x + 3) = 0 \), which simplifies to \( x^2 + x - 6 = 0 \).

Solve the quadratic equation \( 3x^2 + 6x - 9 = 0 \) using the quadratic formula. The positive root is \( x = 1 \).

From the properties of roots, \( b = -(r_1 + r_2) = -8 \) and \( c = r_1 cdot r_2 = 15 \). Therefore, \( b + c = -8 + 15 = 7 \).

Factorize the equation to \( (x-2)(x-5) = 0 \). The roots are \( 2 \) and \( 5 \). The difference between the larger root and the smaller root is \( 5 - 2 = 3 \).

Rearrange the equation to \( x^2 + (k+5)x - 25 = 0 \). For the equation to have only one solution, the discriminant must be zero: \( (k+5)^2 - 4(-25) = 0 \). Solving gives \( k = -15 \) or \( k = 5 \).

Rearrange the equation to \( 2x^2 - 14x + 15 = 0 \). The product of the solutions is given by \( c/a \), where \( a = 2 \) and \( c = 15 \), resulting in \( 15/2 \).