ORIGINAL WORKED EXAMPLE
Two roots, including a negative one
Solve 2x²−5x−3=0.
- Factor as (2x+1)(x−3)=0.
- Set each factor to zero: 2x+1=0 or x−3=0.
- x=−1/2 or x=3. Both satisfy the original polynomial.
Answer −1/2 and 3
GRE · Algebra · TOPIC LESSON
Quadratic equations and inequalities
Connect factors, roots, square structure and sign intervals without importing unnecessary calculus.
BY THE END, YOU CAN
- Solve quadratic equations with factoring and the quadratic formula.
- Use discriminants and completed squares.
- Solve quadratic inequalities and reject extraneous solutions.
BEFORE YOU START
- Linear equations
- Factoring and exponent laws
Original educational lesson. Work through the examples before revealing the practice answers. This original lesson supports current GRE General Test study; it does not reproduce official questions, certify a writing score or predict test content.
LEARN THROUGH ORIGINAL PRACTICE
Original GRE-style worked examples and practice
All teaching passages, writing prompts, examples and exercises in this lesson are original. No official ETS questions, prompts or explanations are reproduced. Current structure and observed public sample metadata are separate from this original teaching; no official score or question probability is implied.
01 / UNDERSTAND & APPLY
Identify the quadratic and its requested feature
A quadratic expression has form ax²+bx+c with a≠0. Its equation ax²+bx+c=0 asks for roots; its inequality asks where the expression has a specified sign. Its graph is a parabola, opening upward for a>0 and downward for a<0. These are related views but answer different questions.
Put an equation equal to zero before using the zero-product property. A product equal to a nonzero number does not allow setting each factor equal to that number.
02 / UNDERSTAND & APPLY
A zero product gives separate linear cases
If uv=0 for real u,v, at least one factor is zero. Factoring a quadratic turns a quadratic equation into linear equations. If a factor is repeated, it gives one distinct root with multiplicity two, not two different values.
Do not divide by a variable factor and silently lose a possible root. x²=5x becomes x(x−5)=0, so x=0 or x=5. Dividing immediately by x keeps only 5.
03 / UNDERSTAND & APPLY
The quadratic formula covers nonfactorable cases
For ax²+bx+c=0, x=[−b±√(b²−4ac)]/(2a). The denominator 2a divides the entire numerator. The discriminant D=b²−4ac determines real roots: two distinct if D>0, one repeated if D=0, none if D<0.
A negative discriminant does not mean a negative root; it means no real roots. A positive discriminant can produce two positive, two negative, or opposite-sign roots. Retain exact radicals unless an approximation is requested.
ORIGINAL WORKED EXAMPLE
Exact irrational roots
Solve x²−4x+1=0.
- Here a=1, b=−4, c=1, so D=16−4=12.
- x=(4±√12)/2=(4±2√3)/2.
- The two solutions are 2−√3 and 2+√3.
Answer 2−√3 and 2+√3
04 / UNDERSTAND & APPLY
A square exposes the turning point and bounds
Complete the square by creating x²+2hx+h²=(x+h)². For x²−6x+11, add and subtract 9 to get (x−3)²+2. A real square is nonnegative, so this expression is at least 2, with equality at x=3.
For a(x−h)²+k, the vertex is (h,k); a>0 gives minimum k and a<0 maximum k over all real x. A restricted domain may exclude the vertex, so a claimed attained minimum must check the permitted inputs.
ORIGINAL WORKED EXAMPLE
Minimum without calculus
Find the minimum of 2x²+8x+13 for real x.
- Factor 2 from the quadratic and linear terms: 2(x²+4x)+13.
- Complete the square: 2[(x+2)²−4]+13=2(x+2)²+5.
- The square reaches zero at x=−2, so the minimum is 5.
Answer 5 at x=−2
05 / UNDERSTAND & APPLY
Roots split the line into sign regions
For (x−r)(x−s) with r<s, the product is positive outside the roots and negative between them, since the two factors have equal signs outside and opposite signs inside. Multiplying by a negative leading coefficient reverses all signs.
For ≤ or ≥, include roots where the expression is zero; for strict inequalities, exclude them. A repeated root does not change the sign across that root. Test one point in each interval rather than assuming every root flips sign.
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ORIGINAL WORKED EXAMPLE
Solve a quadratic inequality
Solve x²−x−6≤0.
- Factor: (x−3)(x+2)≤0. The roots are −2 and 3.
- The leading coefficient is positive, so the product is negative between the roots and positive outside.
- Include both roots because equality is allowed.
Answer [−2,3]
06 / UNDERSTAND & APPLY
Check solutions after operations that can add roots
Squaring both sides can introduce extraneous solutions because opposite numbers have the same square. Clearing denominators excludes values where an original denominator vanishes. Solve the resulting polynomial, then substitute every candidate in the original equation.
For √(x+2)=x, x must first be nonnegative because the left side is. Squaring produces x+2=x², but a negative algebraic candidate cannot satisfy the original sign requirement.
ORIGINAL WORKED EXAMPLE
Reject an extraneous root
Solve √(x+6)=x over the reals.
- The original equation requires x≥0. Square to obtain x+6=x².
- Factor x²−x−6=(x−3)(x+2)=0, giving 3 and −2.
- x=−2 fails the nonnegative right-side requirement. x=3 checks because √9=3.
Answer x=3
07 / UNDERSTAND & APPLY
Use sums and products without solving every root
If a quadratic has roots r and s, factoring a(x−r)(x−s) shows r+s=−b/a and rs=c/a. These relations can answer questions about roots without computing radicals. They describe roots with multiplicity, and a question about real roots still needs a real-root check.
A zero product c=0 guarantees a root at zero. A negative product of real roots means opposite signs. A positive product means same sign, with the sum distinguishing positive from negative when both roots are real.
08 / UNDERSTAND & APPLY
Choose a method and answer the requested question
All examples and practice in this lesson are original. For quantitative comparison (QC), use A if Quantity A is greater, B if Quantity B is greater, C if they are equal, and D if the relationship cannot be determined from the information given. A, B or C must hold for every permitted case. To establish D, give two allowed cases producing different relationships; an inability to calculate is not itself evidence for D.
A single-answer question requires one choice. For multiple-answer questions, obey the stated selection count or select every correct option if instructed. Test each option separately; do not stop after finding one. Multiple-answer questions receive credit only when the complete correct set is selected, with no incorrect choices. Numeric entry requires the requested value, with units handled in the calculation; round only when the question directs you to. A single numeric-entry box accepts an integer or decimal; a pair of fraction boxes accepts an integer numerator and a nonzero integer denominator. Fractions need not be reduced, but reducing makes checks easier. Do not enter a percent sign when the requested number is a percentage. Preserve exact fractions until the last step.
A calculator can check arithmetic but cannot decide the domain, interpretation or denominator for you. Estimate sign and size first, enter parentheses carefully, and compare the result with the original conditions. Cancel factors or exploit structure when doing so is faster and more reliable than decimal entry.
09 / UNDERSTAND & APPLY
Original practice: roots, ranges and signs
For multiple-answer questions, select the complete correct set.
ORIGINAL PRACTICE · 1
Select all that apply: Solve x²=7x. A −7; B 0; C 1; D 7; E 49.
Reveal answer & reasoning
B and D
Move all terms to one side: x(x−7)=0. Dividing by x would wrongly discard the zero solution.
ORIGINAL PRACTICE · 2
Numeric entry: How many distinct real roots does x²+6x+9=0 have?
Reveal answer & reasoning
1
The polynomial is (x+3)², with repeated root −3. Multiplicity two is not two distinct real values.
ORIGINAL PRACTICE · 3
Single answer: Which equation has no real solution? A x²=4; B x²−2=0; C x²+4=0; D (x−2)²=0; E x²−x=0.
Reveal answer & reasoning
C
A real square cannot equal −4. All the other equations have at least one real root.
ORIGINAL PRACTICE · 4
QC: x is real. Quantity A: (x−5)²+3. Quantity B: 2.
Reveal answer & reasoning
A
The square is nonnegative, so A≥3>2 for every x. This is a universal bound, not a conclusion from sample inputs.
ORIGINAL PRACTICE · 5
Numeric entry: The roots of 3x²−12x+5=0 are r and s. Find r+s.
Reveal answer & reasoning
4
The sum is −b/a=12/3=4. The product would instead be 5/3.
ORIGINAL PRACTICE · 6
Select all that apply: Which satisfy (x−1)(x−4)>0? A −2; B 1; C 2; D 4; E 6.
Reveal answer & reasoning
A and E
The product is positive for x<1 or x>4. Roots 1 and 4 give zero and are excluded; 2 lies in the negative interval.
ORIGINAL PRACTICE · 7
Numeric entry: What is the minimum value of x²+10x+28 for real x?
Reveal answer & reasoning
3
Complete the square: (x+5)²+3, minimized at x=−5. The constant term 28 is the value at zero, not the minimum.
ORIGINAL PRACTICE · 8
QC: x²=9. Quantity A: x. Quantity B: 0.
Reveal answer & reasoning
D
The equation permits x=−3 and x=3, giving opposite comparisons. Replacing the equation by x=√9 would lose a root.
KEEP THE BIG IDEAS
Quick revision
- Set a product equal to zero before splitting its factors.
- The discriminant counts real roots; it does not determine their signs alone.
- Completed squares give bounds without calculus.
- Test sign intervals and check candidates in the original equation.
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References & scope
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