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This is the second of our many quizzes on discriminants. Click on this to access more such competitive quizzes on quadratic and other STEM subjects.

Discriminants Quiz 2

mathematics

 Find nature of roots of  $ - {{A}}{{{x}}^2} + 2{{Ax + 20 = 0}}$

 From the given equation, determine the value of D, then determine the possible value range of D, with the help of possible value range estimate the nature of roots.

  $\text{As, }D = {b^2} - 4ac \\D  =  (2A)^2 - 4 \times A \times 20 \\{{D =  4A}}\left( {{{A}} - {{20}}} \right)\\ \text{ Which is negative if }{{A}} \in \left( {0,20} \right)\\ \text{ And positive if }{{A}} \in \left( { - \infty ,0} \right){{U}}\left( {20,\infty } \right)$

 If ${{k}} \in $ integer and roots of ${{{x}}^2} + 2{{x + k = 0}}$ are rational then how many values ${{k}}$ can take?

 For rational roots, discriminant should be a perfect square, From the expression of D determine the possible value of k.

  $\text{As, }{{D  =  }}{b^2} - 4ac \\  {{D  =  (2}}{{{)}}^2} - 4 \times 1 \times k \\{{D =  4}}\left( {1 - {{k}}} \right)\\\text{W.K.T, for rational roots “D” should be perfect square,}\\ \therefore \,{{k  =  1,}}\,{{0,}}\, - 3,\, - 8,\, - 15,\, - 24.......$

 If roots of ${{{x}}^2} + 4{{x + }}{{{k}}^2}{{ = 0}}$ are rational then how many integral values ${{k}}$ can take?

 For rational roots, discriminant should be a perfect square. From the value of D determine the possible value range of k, such that the value of D should be a perfect square.

 $\text{As, }{{D  =  }}{b^2} - 4ac \\  D  =  {4^2} - 4 \times 1 \times {k^2} \\{{ D =  4}}\left( {4 - {{{k}}^2}} \right)\\\text{ As we know that, for rational roots,} \\ \text{We have to make “D” perfect square }\\\therefore \,{{k  =  }}\,{{0,}}\,2,\, - 2$

 Find the nature of roots of $\left( {{{x}} - 1} \right)\,\left( {{{x}} - 2} \right) + \,\left( {{{x}} - 2} \right)\,\left( {{{x}} - 3} \right) + \left( {{{x}} - 3} \right)\,\left( {{{x}} - 1} \right)\, = \,0$.

 Solve the expression and convert it in general form of Quadratic equation, then determine the value of D, with the help of value of D, estimate the nature of roots.

 $\text{Above equation can be written as, }\\ \left( {{{x}} - 1} \right)\,\left( {{{x}} - 2} \right) + \,\left( {{{x}} - 2} \right)\,\left( {{{x}} - 3} \right) + \left( {{{x}} - 3} \right)\,\left( {{{x}} - 1} \right)\, = \,0 \\  [{x^2} - 3x + 2] + [{x^2} - 5x + 6] + [{x^2} - 4x + 3] = 0 \\  {x^2} - 12x + 11 = 0 \\ \text{ Value of “D” will be, }\\{{D  =  }}{\left( { - {{12}}} \right)^2} - \,4\left( 3 \right)\left( {11} \right) \Rightarrow  12\\ \text{Hence, roots are real and distinct.}$

 If roots of ${{{x}}^2} + {{Ax + 1 = 0}}$ are real then roots of ${{A}}{{{x}}^2} + 2{{x + A = 0}}$ are 

 From given expression determine the value of D, and then estimate the nature of roots.ṣ

 $\text{As, }{{D  =  }}{b^2} - 4ac \\\Rightarrow {{D  = }}{{{A}}^2} - 4 \geqslant {{0 }}\\\therefore \,{{{A}}^2} \geqslant 4\\ \text{Now, equating the above equation, }{{D  =  4}} - 4{{{A}}^2}\\{D{ =  4}}\left( {1 - {{{A}}^2}} \right)\text{ Now }{{{A}}^2} \geqslant 4\\\therefore \,{{ 4}}\left( {1 - {{{A}}^2}} \right) < 0\\ \text{Hence roots are imaginary.}$