CBSE Questions for Class 10 Maths Quadratic Equations Quiz 5 - MCQExams.com

Given quadratic equation is $$\sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0 $$   

Comparing it with the standard form of equation $$ax^2+bx+c$$ we get $$a=7,b=6,c=13\sqrt 7$$ 
Therefore, $$x= \dfrac { -b\pm \sqrt { b^ 2-4ac }  }{ 2a } $$
$$=   \dfrac {  -(-6)\pm \sqrt { (-6)^{ 2 }-4\times \sqrt { 7 } \times \left( -13\sqrt { 7 }  \right)  }  }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6\pm \sqrt { 36+364 }  }{ 2\sqrt { 7 }  } $$
$$= \dfrac {  6\pm \sqrt { 400 }  }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6\pm 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac {  6- 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\- 14 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ -7 }{ \sqrt { 7 }  } $$
$$=\dfrac { \\ -7\times \sqrt { 7 }  }{ \sqrt { 7 } \times \sqrt { 7 }  } $$
$$=-\sqrt 7$$
$$=\dfrac {  6+ 20 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ 26 }{ 2\sqrt { 7 }  } $$
$$=\dfrac { \\ 13 }{ \sqrt { 7 }  } $$

The quadratic equation is $$2x^2 - 2\sqrt 2x + 1 = 0$$.
Comparing it with standard form of quadratic equation $$ax^2+bx+c=0$$,we get $$a=2,b=- 2\sqrt 2,c=1$$ Therefore the roots of the quadratic equation are,
$$x= \dfrac { -b\pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
  $$= \dfrac { -\left( -2\sqrt { 2 }  \right) \pm \sqrt { \left( -2\sqrt { 2 }  \right) ^{ 2 }-4\times 2\times 1 }  }{ 2\times 2 } $$
  $$=\dfrac { \left( 2\sqrt { 2 }  \right) \pm \sqrt { 8-8 }  }{ 4 } $$
  $$=\dfrac { 2\sqrt { 2 } \pm 0 }{ 4 } $$
  $$=\dfrac { 2\sqrt { 2 }  }{ 4 } $$  
  $$=\dfrac { \sqrt { 2 }  }{ 2 } $$ 
  $$=\dfrac{ 1 }{ \sqrt { 2 }  } $$ 

Given, $$\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 4$$
$$\Rightarrow \dfrac { x^{ 2 }+(x-1)^{ 2 } }{ x(x-1) } =4$$
$$\Rightarrow x^2+x^2-2x+1=4x^2-4x$$
$$\Rightarrow 2x^2-4x^2-2x+4x+1=0$$
$$\Rightarrow -2x^2+2x+1=0$$
$$\Rightarrow 2x^2-2x-1=0$$
Comparing $$2x^2-2x-1=0$$ with $$ax^2+bx+c=0$$, we get $$a=2, b=-2. c=-1$$
Now the roots are $$=$$ $$\dfrac { -b \pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
$$=$$ $$\dfrac { -(-2)\pm \sqrt { (-2)^ 2-4\times 2\times (-1) }  }{ 2\times 1 }$$
$$=$$ $$ \dfrac { 2\pm \sqrt { 4+8 }  }{ 2 }$$
$$=$$ $$\dfrac { 2\pm \sqrt { 12 }  }{ 2 }$$
$$=$$ $$\dfrac { 2\pm 2\sqrt { 3 }  }{ 2 }$$
$$=$$ $$1\pm \sqrt 3$$

Given, $$x+\dfrac{1}{x}=3$$
$$\Rightarrow \dfrac{x+1}{x}=3$$
$$\Rightarrow x^2-3x+1=0$$
Comparing $$x^2-3x+1=0$$ with $$ax^2+bx+c=0$$, we get $$a=1, b=3. c=14$$
Now the roots are $$=$$ $$\dfrac { -b \pm \sqrt { b^{ 2 }-4ac }  }{ 2a } $$
$$=$$ $$\dfrac { -(-3)\pm \sqrt { (3)^ 2-4\times 1\times 1 }  }{ 2\times 1 } $$
$$=$$ $$\dfrac { 3\pm \sqrt { 9-4 }  }{ 2 }$$
$$=$$ $$ \dfrac { 3\pm \sqrt { 5 }  }{ 2 }$$
$$=$$ $$ \dfrac { 3+ \sqrt { 5 }  }{ 2 }$$ or $$\dfrac { 3- \sqrt { 5 }  }{ 2 }$$