Mathematics · Polynomials · NCERT Exercises

Exercise 2.2

Complete, independently verified solutions for NCERT Exercise 2.2.

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Mathematics · Chapter 2 · NCERT Exercises

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Question 1

quadratic polynomials · zeroes and coefficients

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

  1. x2−2x−8x^2-2x-8
  2. 4s2−4s+14s^2-4s+1
  3. 6x2−3−7x6x^2-3-7x
  4. 4u2+8u4u^2+8u
  5. t2−15t^2-15
  6. 3x2−x−43x^2-x-4

Solution

  1. (x−4)(x+2);4,−2;α+β=2=−ba, αβ=−8=ca(x-4)(x+2);\quad 4,-2;\quad \alpha+\beta=2=-\frac ba,\ \alpha\beta=-8=\frac ca
  2. (2s−1)2;12,12;α+β=1=−ba, αβ=14=ca(2s-1)^2;\quad \frac12,\frac12;\quad \alpha+\beta=1=-\frac ba,\ \alpha\beta=\frac14=\frac ca
  3. (3x+1)(2x−3);−13,32;α+β=76=−ba, αβ=−12=ca(3x+1)(2x-3);\quad -\frac13,\frac32;\quad \alpha+\beta=\frac76=-\frac ba,\ \alpha\beta=-\frac12=\frac ca
  4. 4u(u+2);0,−2;α+β=−2=−ba, αβ=0=ca4u(u+2);\quad 0,-2;\quad \alpha+\beta=-2=-\frac ba,\ \alpha\beta=0=\frac ca
  5. (t−15)(t+15);15,−15;α+β=0=−ba, αβ=−15=ca(t-\sqrt{15})(t+\sqrt{15});\quad \sqrt{15},-\sqrt{15};\quad \alpha+\beta=0=-\frac ba,\ \alpha\beta=-15=\frac ca
  6. (3x−4)(x+1);43,−1;α+β=13=−ba, αβ=−43=ca(3x-4)(x+1);\quad \frac43,-1;\quad \alpha+\beta=\frac13=-\frac ba,\ \alpha\beta=-\frac43=\frac ca

Question

Question 2

forming a quadratic polynomial · sum and product of zeroes

Find a quadratic polynomial in each case when the given numbers are respectively the sum and product of its zeroes.

  1. 14, −1\frac14,\ -1
  2. 2, 13\sqrt2,\ \frac13
  3. 0, 50,\ \sqrt5
  4. 1, 11,\ 1
  5. −14, 14-\frac14,\ \frac14
  6. 4, 14,\ 1

Solution

For sum S and product P, one suitable polynomial is x² − Sx + P. Multiplying by a non-zero constant gives an equivalent answer.

  1. 4x2−x−44x^2-x-4
  2. 3x2−32x+13x^2-3\sqrt2x+1
  3. x2+5x^2+\sqrt5
  4. x2−x+1x^2-x+1
  5. 4x2+x+14x^2+x+1
  6. x2−4x+1x^2-4x+1