EAPCET Mathematics · Intermediate second year · Paper IIA

Quadratic Expressions

Quadratic Expressions studies the roots of quadratic equations, their nature, the relations with coefficients, and the sign and extreme values of a quadratic expression.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIA
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • Roots: x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
  • α+β=−ba\alpha + \beta = -\dfrac{b}{a} and αβ=ca\alpha\beta = \dfrac{c}{a}
  • Δ=b2−4ac>0\Delta = b^{2} - 4ac > 0: real distinct roots; =0= 0: equal roots; <0< 0: complex conjugate roots
  • If a>0a > 0, ax2+bx+cax^{2} + bx + c has minimum value 4ac−b24a\dfrac{4ac - b^{2}}{4a} at x=−b2ax = -\dfrac{b}{2a}
  • ax2+bx+c>0ax^{2} + bx + c > 0 for every real xx iff a>0a > 0 and Δ<0\Delta < 0
  • (α−β)2=(α+β)2−4αβ(\alpha - \beta)^{2} = (\alpha + \beta)^{2} - 4\alpha\beta, so ∣α−β∣=Δ∣a∣|\alpha - \beta| = \dfrac{\sqrt{\Delta}}{|a|}
  • Quadratic with roots α\alpha and β\beta: x2−(α+β)x+αβ=0x^{2} - (\alpha + \beta)x + \alpha\beta = 0
  • ax2+bx+c=0ax^{2} + bx + c = 0 and dx2+ex+f=0dx^{2} + ex + f = 0 have one common root iff (af−cd)2=(bd−ae)(ce−bf)(af - cd)^{2} = (bd - ae)(ce - bf)
  • Both roots common: ad=be=cf\dfrac{a}{d} = \dfrac{b}{e} = \dfrac{c}{f}
  • Both roots greater than kk: Δ≥0\Delta \ge 0, a⋅f(k)>0a \cdot f(k) > 0 and −b2a>k-\dfrac{b}{2a} > k (reverse the last for both less than kk)
  • kk lies between the roots: a⋅f(k)<0a \cdot f(k) < 0
  • Both roots between k1k_{1} and k2k_{2}: Δ≥0\Delta \ge 0, a⋅f(k1)>0a \cdot f(k_{1}) > 0, a⋅f(k2)>0a \cdot f(k_{2}) > 0 and k1<−b2a<k2k_{1} < -\dfrac{b}{2a} < k_{2}
  • Exactly one root between k1k_{1} and k2k_{2}: f(k1)⋅f(k2)<0f(k_{1}) \cdot f(k_{2}) < 0

What this chapter covers

  • Roots and the discriminant
  • Relations between roots and coefficients
  • Common roots
  • Sign of a quadratic expression
  • Maximum and minimum values
  • Quadratic inequations

How it is usually asked

  • Form an equation whose roots are given functions of α and β
  • Find the range of a parameter for real or positive roots
  • Find a maximum or minimum value
  • Find the condition for a common root

Understanding Quadratic Expressions

A quadratic equation ax² + bx + c = 0 with a ≠ 0 has roots [−b ± √(b² − 4ac)]/2a. The discriminant Δ = b² − 4ac decides their nature: positive gives two distinct real roots, zero gives equal roots, and negative gives a pair of complex conjugate roots. If the coefficients are rational and Δ is a perfect square, the roots are rational.

Without solving, the roots α and β satisfy α + β = −b/a and αβ = c/a. Any symmetric expression in α and β, such as α² + β² = (α + β)² − 2αβ, can be found from these two. An equation with given roots is x² − (sum of roots)x + (product of roots) = 0, which is how questions about transformed roots are answered.

As a function, y = ax² + bx + c is a parabola. When a > 0 it has a minimum value (4ac − b²)/4a at x = −b/2a; when a < 0 that is a maximum. The expression keeps the sign of a for all x when Δ < 0, and has the sign of a outside the roots and the opposite sign between them when Δ > 0. EAPCET uses these facts for parameter ranges and inequalities.

Worked example

EAPCET-style question

If α\alpha and β\beta are the roots of x2−5x+3=0x^{2} - 5x + 3 = 0, the equation whose roots are α2\alpha^{2} and β2\beta^{2} is

  1. (1)x2−19x+9=0x^{2} - 19x + 9 = 0
  2. (2)x2+19x+9=0x^{2} + 19x + 9 = 0
  3. (3)x2−25x+9=0x^{2} - 25x + 9 = 0
  4. (4)x2−19x+3=0x^{2} - 19x + 3 = 0

Solution

  1. α+β=5\alpha + \beta = 5 and αβ=3\alpha\beta = 3.
  2. Sum of new roots: α2+β2=(α+β)2−2αβ=25−6=19\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = 25 - 6 = 19.
  3. Product of new roots: α2β2=(αβ)2=9\alpha^{2}\beta^{2} = (\alpha\beta)^{2} = 9.
  4. The equation is x2−19x+9=0x^{2} - 19x + 9 = 0.

Answer: x2−19x+9=0x^{2} - 19x + 9 = 0

Shortcuts and results to remember

  • Both roots positive: Δ≥0\Delta \ge 0, −ba>0-\tfrac{b}{a} > 0, ca>0\tfrac{c}{a} > 0; both negative: Δ≥0\Delta \ge 0, −ba<0-\tfrac{b}{a} < 0, ca>0\tfrac{c}{a} > 0; opposite signs: ca<0\tfrac{c}{a} < 0.
  • Range of y=linear or quadraticquadraticy = \dfrac{\text{linear or quadratic}}{\text{quadratic}}: write it as a quadratic in xx with yy as a parameter and demand Δ≥0\Delta \ge 0.
  • For a<0a < 0, ax2+bx+cax^{2} + bx + c has maximum value 4ac−b24a\dfrac{4ac - b^{2}}{4a}, so its range is (−∞,4ac−b24a]\left(-\infty, \dfrac{4ac - b^{2}}{4a}\right].
  • Useful factorisation: a4+a2+1=(a2+a+1)(a2−a+1)a^{4} + a^{2} + 1 = (a^{2} + a + 1)(a^{2} - a + 1).

More practice questions

Practice question 1

If x2−kx−21=0x^{2} - kx - 21 = 0 and x2−3kx+35=0x^{2} - 3kx + 35 = 0 have a common root, then kk equals

  1. (1)±4
  2. (2)±2
  3. (3)±3
  4. (4)±5

Solution

  1. Subtract the equations: 2kx−56=02kx - 56 = 0, so the common root is x=28kx = \dfrac{28}{k}
  2. Substitute in the first: 784k2−28−21=0\dfrac{784}{k^{2}} - 28 - 21 = 0
  3. 784k2=49\dfrac{784}{k^{2}} = 49, so k2=16k^{2} = 16 and k=±4k = \pm 4
  4. Check k=4k = 4: (x−7)(x+3)(x - 7)(x + 3) and (x−5)(x−7)(x - 5)(x - 7) share x=7x = 7

Answer: ±4

Practice question 2

The number 5 lies between the roots of x2−2kx+k2−4=0x^{2} - 2kx + k^{2} - 4 = 0 exactly when

  1. (1)3<k<73 < k < 7
  2. (2)k<3k < 3
  3. (3)k>7k > 7
  4. (4)−2<k<2-2 < k < 2

Solution

  1. Here a=1>0a = 1 > 0, so the condition is f(5)<0f(5) < 0
  2. 25−10k+k2−4<025 - 10k + k^{2} - 4 < 0, i.e. k2−10k+21<0k^{2} - 10k + 21 < 0
  3. (k−3)(k−7)<0(k - 3)(k - 7) < 0 gives 3<k<73 < k < 7
  4. Check: the roots are k±2k \pm 2, and k−2<5<k+2k - 2 < 5 < k + 2 gives the same interval

Answer: 3<k<73 < k < 7

Common mistakes to avoid

  • Writing α + β = b/a; the sum of roots is −b/a.
  • Using α² + β² = (α + β)² and forgetting to subtract 2αβ.
  • Using Δ > 0 when the question asks for real roots; real roots include equal roots, so the condition is Δ ≥ 0.
  • Taking the solution of a quadratic inequality between the roots when a > 0 and the expression must be positive; for a > 0 it is positive outside the roots.

How to prepare it

Learn the sign scheme for quadratic inequalities: outside the roots when a > 0, between them when a < 0.

Once the formulas feel familiar, test them the way the exam does: practise Quadratic Expressions questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Quadratic Expressions chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Quadratic Expressions part of the EAPCET Mathematics syllabus?
Yes. Quadratic Expressions is taught in Intermediate second year · Paper IIA Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Quadratic Expressions for EAPCET?
Learn the sign scheme for quadratic inequalities: outside the roots when a > 0, between them when a < 0.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Quadratic Expressions?
Writing α + β = b/a; the sum of roots is −b/a. Using α² + β² = (α + β)² and forgetting to subtract 2αβ. Using Δ > 0 when the question asks for real roots; real roots include equal roots, so the condition is Δ ≥ 0. Taking the solution of a quadratic inequality between the roots when a > 0 and the expression must be positive; for a > 0 it is positive outside the roots.