SAT equivalent expressions and quadratics: review and practice

• Rewrite an expression in an equivalent form by combining terms, factoring or using exponent rules.
• Solve quadratic equations by factoring or with the quadratic formula.
• Use the discriminant to tell whether a quadratic has two, one or no real solutions.
• Solve a system where one equation is a line and the other is a parabola.
• Check answers in the original equation, especially after squaring both sides or when a denominator could be zero.

• Adding exponents when you should multiply, or the reverse. x⁵ · x³ = x⁸, but (x⁵)³ = x¹⁵.
• Losing a sign in the quadratic formula. For x² − 5x − 3 = 0, b is −5, so −b is +5 and b² − 4ac is 25 + 12 = 37.
• Reading a, b and c before setting the equation to zero. For x² = 4x + 5, rewrite it as x² − 4x − 5 = 0 first.
• Dropping the negative square root. If x² = 49, x can be 7 or −7. A question asking for the smallest solution is counting on you to forget −7.
• Canceling across addition. (x + 6)/6 is not x + 1. You can only cancel factors: (6x + 12)/6 = x + 2.
• Keeping a solution that breaks a condition. If the problem says p is negative and your algebra gives p = 3 or p = −5, the answer is −5.

1. Question 1 · Easy

   Which expression is equivalent to (5y² + 14y) + (2y² − 14y)?

   • A7y² + 28y
   • B7y²
   • C7y⁴
   • D3y²
   Show the answer

   B. The quadratic terms add to 7y², and 14y − 14y = 0, so the expression equals 7y². The other choices add exponents when adding terms, subtract the quadratic coefficients, or change the negative linear term to a positive term.

2. Question 2 · Easy

   Which expression is equivalent to x²¹ · x⁷ for all positive values of x?

   • Ax¹⁴
   • Bx¹⁴⁷
   • Cx²⁸
   • Dx³
   Show the answer

   C. The product is x^(21 + 7) = x²⁸ because multiplying powers with a common base adds their exponents. The exponents 3, 14, and 147 result from dividing, subtracting, and multiplying the given exponents, respectively.

3. Question 3 · Medium

   What is the greatest solution of 2x² − 3x − 8 = 0?

   • A(3 − sqrt{73})/2
   • B(3 − sqrt{73})/4
   • C(3 + sqrt{73})/4
   • D(3 + sqrt{73})/2
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   C. The quadratic formula gives x = (3 ± sqrt{9 + 64})/4 = (3 ± sqrt{73})/4, so the greatest solution is (3 + sqrt{73})/4. The other choices use the smaller branch, omit the factor of 2 in the denominator, or make both errors.

4. Question 4 · Medium

   2x² + (3p + 6)x + 18 = 0 In the given equation, p is a negative constant. The equation has exactly one real solution. What is the value of p?

   • A−6
   • B−4
   • C−2
   • D2
   Show the answer

   A. For a = 2, b = 3p + 6, and c = 18, the repeated-root condition gives (3p + 6)² = 144, so 3p + 6 = 12 or 3p + 6 = −12. These give p = 2 or p = −6, and the negative-constant condition selects p = −6. The other choices omit the shift in the linear coefficient, set that coefficient itself to zero, or keep the positive parameter branch.

5. Question 5 · Hard

   y = 2x² − 19 y = px − 27 In the system shown, p is a negative constant. The graphs of the equations intersect at exactly one point (a, b) in the xy-plane. What is the value of a?

   • A−11
   • B−8
   • C−4
   • D−2
   Show the answer

   D. Eliminating y gives 2x² − px + 8 = 0, so one intersection requires p² − 64 = 0; since p < 0, p = −8 and a = p/4 = −2. Substitution gives the point (−2, −11); the other choices report the y coordinate, the parameter p, or twice the repeated root.

6. Question 6 · Hard

   Which of the following expressions has a factor of x + 3b, where b is a positive integer constant?

   • A4x² + 5x + 15b
   • B4x² + 35x + 15b
   • C4x² + 23x + 15b
   • D4x² + 29x + 15b
   Show the answer

   D. Since b > 0, the other factor must be 4x + 5, and (x + 3b)(4x + 5) = 4x² + (12b + 5)x + 15b. The middle coefficient 29 gives b = 2, whereas coefficients 5, 23, and 35 give b = 0, 3/2, and 5/2, respectively; only 2 is a positive integer, so 4x² + 29x + 15b is the correct expression.

Type the quadratic as y = x² − 5x − 3 and click where the parabola crosses the x-axis; Desmos marks each root with a gray point and shows its coordinates. For a line–parabola system, type both equations and click the intersections. When an unknown constant decides the number of solutions, type it as k, add the slider and drag until the line just touches the parabola, then confirm the exact value with b² − 4ac = 0.

Ashore's SAT course starts with a 15-minute check that finds the question types costing you the most points, then gives you lessons and practice for each.

• Linear Equations and Inequalities
• Linear Functions and Systems
• Ratios, Rates, Percentages, and Units
• One- and Two-Variable Data
• Probability and Statistical Inference
• Nonlinear Functions and Exponentials
• Lines, Angles, and Triangles
• Right Triangles and Trigonometry
• Circles, Area, and Volume
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