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## Algebra: Quadratic Equations Test-3

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Question 1 |

Which of the following is a quadratic equation?

$ \displaystyle {{x}^{\frac{1}{2}}}+2x+3=0$ | |

$ \displaystyle \left( x-1 \right)\,\left( x+4 \right)={{x}^{2}}+1$ | |

$ \displaystyle {{x}^{4}}-3x+5=0$ | |

$ \displaystyle \left( 2x+1 \right)\,\left( 3x-4 \right)\,=2{{x}^{2}}+3$ |

Question 1 Explanation:

By observation the highest power of x is 2 only in option (d)

Question 2 |

Which one of the following is a factor of

$ \displaystyle {{x}^{3}}-19x+30$

$ \displaystyle {{x}^{3}}-19x+30$

$ \displaystyle x-2$ | |

$ \displaystyle x+2$ | |

$ \displaystyle x-1$ | |

$ \displaystyle x+1$ |

Question 3 |

The solution of the equation

$ \displaystyle \sqrt{25-{{x}^{2}}}=x-1$ are

$ \displaystyle \sqrt{25-{{x}^{2}}}=x-1$ are

$ \displaystyle x=3\,\,\,and\,\,x=4$ | |

$ \displaystyle x=5\,\,\,\,and\,\,\,\,x=1$ | |

$ \displaystyle x=-3\,\,and\,\,x=4$ | |

$ \displaystyle x=4\,\,\,\,and\,\,\,x\ne -3$ |

Question 3 Explanation:

For x =4 , the equation

$latex \sqrt{25-{{x}^{2}}}=x-1$ is satisfied ,

Thus options which can be correct are a, c and d. Now when we put x=-3 we find that both sides are not equal so we can conclude that (d) is the answer)

$latex \sqrt{25-{{x}^{2}}}=x-1$ is satisfied ,

Thus options which can be correct are a, c and d. Now when we put x=-3 we find that both sides are not equal so we can conclude that (d) is the answer)

Question 4 |

Of the following quadratic equations

which is the one whose roots are 2 and −15?

which is the one whose roots are 2 and −15?

$ \displaystyle {{x}^{2}}-2x+15=0$ | |

$ \displaystyle {{x}^{2}}+15x-2=0$ | |

$ \displaystyle {{x}^{2}}+13x-30=0$ | |

$ \displaystyle {{x}^{2}}-30=0$ |

Question 4 Explanation:

The factors are x-2 and x+15.

Thus the quadratic equation is

$ \begin{array}{l}(x-2)(x+15)\\={{x}^{2}}+13x-30\end{array}$

Thus the quadratic equation is

$ \begin{array}{l}(x-2)(x+15)\\={{x}^{2}}+13x-30\end{array}$

Question 5 |

If $ \displaystyle \begin{array}{l}{{\left( a-3 \right)}^{2}}+{{\left( b-4 \right)}^{2}}+{{\left( c-9 \right)}^{2}}=0,\,\,\\then\,\,the\,\,value\,\,of\,\,\sqrt{a+b+c}\end{array}$ is

-4 | |

+4 | |

±4 | |

±2 |

Question 5 Explanation:

Since the sum of three squares can only be equal to 0

if the terms are individually equal to 0

we can conclude that a=3,b=4,c=9.

The sum of a, b, c = 16.

Thus, $ \sqrt{a+b+c}=\sqrt{16}=\pm 4$

if the terms are individually equal to 0

we can conclude that a=3,b=4,c=9.

The sum of a, b, c = 16.

Thus, $ \sqrt{a+b+c}=\sqrt{16}=\pm 4$

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