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## Algebra: Quadratic Equations Test-4

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Question 1 |

The expression $ \displaystyle {{x}^{4}}-2{{x}^{2}}+k$

will be a perfect square

when the value of K is

will be a perfect square

when the value of K is

2 | |

1 | |

âˆ’1 | |

âˆ’2 |

Question 1 Explanation:

$ \begin{array}{l}{{x}^{4}}-2{{x}^{2}}+k\\={{x}^{4}}-2{{x}^{2}}+1+k-1\\={{({{x}^{2}}-1)}^{2}}+k-1\\If\,k=1\,\text{then}\,\text{the}\,\text{term}\,\text{will}\,\text{be}\,\text{a}\,\text{perfect}\,\text{square}\text{.}\end{array}$

Question 2 |

if pâˆ’2q=4, then the value of p

^{3}âˆ’8q^{3}âˆ’24pqâˆ’64 is:2 | |

0 | |

3 | |

-1 |

Question 2 Explanation:

$ \displaystyle \begin{array}{l}p-2q=4\\Cubing\,both\,sides,\\=>{{(p-2q)}^{3}}={{4}^{3}}\\=>{{p}^{3}}-3.{{p}^{2}}.2q+3.p.4{{q}^{2}}-8{{q}^{3}}=64\\=>{{p}^{3}}-6pq(p-2q)-8{{q}^{3}}=64\\=>{{p}^{3}}-24pq-8{{q}^{3}}-64=0\end{array}$

Question 3 |

If the expression x

^{2}+x +1 is<br>written in the formÂ $\displaystyle {{\left( x+\frac{1}{2} \right)}^{2}}+{{q}^{2}}$<br>then the possible values of q are$ \displaystyle \pm \frac{1}{3}$ | |

$ \displaystyle \pm \frac{\sqrt{3}}{2}$ | |

$ \displaystyle \pm \frac{2}{\sqrt{3}}$ | |

$ \displaystyle \pm \frac{1}{2}$ |

Question 3 Explanation:

$ \displaystyle \begin{array}{l}{{\left( x+\frac{1}{2} \right)}^{2}}+{{q}^{2}}={{x}^{2}}+x+1\\=>{{x}^{2}}+\frac{1}{4}+x+{{q}^{2}}={{x}^{2}}+x+1\\=>{{q}^{2}}=\frac{3}{4}\\=>q=\pm \sqrt{\frac{3}{4}}=\pm \frac{1}{2}\sqrt{3}\end{array}$

Question 4 |

$ \displaystyle {{a}^{2}}-2a-1=0$

then value of $ \displaystyle {{a}^{2}}+\frac{1}{{{a}^{2}}}+3a-\frac{3}{a}$ is

then value of $ \displaystyle {{a}^{2}}+\frac{1}{{{a}^{2}}}+3a-\frac{3}{a}$ is

25 | |

30 | |

35 | |

40 |

Question 4 Explanation:

$ \displaystyle \begin{array}{l}{{a}^{2}}-2a-1=0\\{{a}^{2}}-1=2a\\on\,dividing\,with\,a\\a-\frac{1}{a}=2\\Now\,\,{{a}^{2}}+\frac{1}{{{a}^{2}}}+3a-\frac{3}{a}\\{{\left( a-\frac{1}{a} \right)}^{2}}+2+3\left( a-\frac{1}{a} \right)\\4+2+3(2)\\4+2+6=12\end{array}$

Question 5 |

If a

^{2}+1=a, then the value of a^{12}+a^{6}+1 is:-3 | |

1 | |

2 | |

3 |

Question 5 Explanation:

\[\begin{align}
& {{a}^{2}}+1=a \\
& =>{{a}^{2}}-a+1=0 \\
& Multiply\,both\,sides\,by\,(a+1), \\
& =>(a+1)({{a}^{2}}-a+1)=0 \\
& =>{{a}^{3}}+1=0 \\
& =>{{a}^{3}}=\,-1 \\
& Thus, \\
& {{a}^{12}}+{{a}^{6}}+1=1+1+1=3 \\
& \\
& Alternate\,solution \\
& {{a}^{2}}+1=a \\
& Cubing\,both\,sides \\
& =>{{a}^{6}}\,+\,1+\,3{{a}^{2}}\,({{a}^{2}}\,+\,1)\,=\,{{a}^{3}} \\
& =>{{a}^{6}}\,+\,1+\,3{{a}^{2}}\,(a)\,=\,{{a}^{3}}\,\,(initial\,condition\,used) \\
& =>{{a}^{6}}\,+\,1+\,3{{a}^{3}}\,=\,{{a}^{3}} \\
& =>{{a}^{6}}\,+\,1+\,2{{a}^{3}}\,\,=\,0 \\
& =>\,\,{{({{a}^{3}}\,+\,1)}^{2}}\,=\,0 \\
& =>\,\,{{a}^{3}}\,\,=\,-1 \\
\end{align}\]
Using this, we can find the value of the expression.
Remember, in this case, only one real root (a=-1) exists whereas the other two roots are imaginary in nature.

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There are 5 questions to complete.

List |

Question 5 Explanation: a2+1=aCubingbothsides,=>a2âˆ’a+1=0=>(a+1)(a2âˆ’a+1)=0=>a3+1=0=>a=âˆ’1âˆ’âˆ’âˆ’âˆš3Thusa12+a6+1=1+1+1=3

This is wrong..

U multiplied by (a+1) and this expression is getting 0 at a=-1, and u have assumed this as solution for quadratic equation.

Actual solution for quadratic equation is +-i.

So answer is -3 for this solution.