## Inquiry

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# Quadratic Equation by Discussion

1. Use the discriminant to find out the count of the roots of following equations:

x^{2}–14x +33=0 |
D=(-14)^{2}-4.1.33=64 |
Two different roots in real numbers |

4x^{2}–5x+1=0 |
D=(-5)^{2}-4.4.1=9 |
Two different roots in real numbers |

x^{2}–10x+25=0 |
D=(-10)^{2}-4.1.25=0 |
One (double) root in real numbers |

12x^{2}–5x-3=0 |
D=(-5)^{2}-4.12.(-3)=169 |
Two different roots in real numbers |

x^{2}–4x+13=0 |
D=(-4)^{2}-4.1.13=–36 |
No solution in real numbers |

x^{2}–14x+49=0 |
D=(-14)^{2}–4.1.49=0 |
One (double) root in real numbers |

x^{2}–6x+25=0 |
D=(-6)^{2}-4.1.25=–64 |
No solution in real numbers |

2. For which „m“ do the quadratic equations have one (double) real root?

^{2}+4x+m = 0

D = 0

b

^{2}-4ac = 0

4

^{2}-4.1.m = 0

4m = 16

m = 4

x

^{2}+4x+4=0

^{2}+(4m-2)x+(4m+1) = 0

D = 0

b

^{2}–4ac = 0

(4m-2)

^{2}–4m(4m+1) = 0

16m

^{2}–16m+4–16m

^{2}–4m = 0

–20m = –4

m = 1/5

x

^{2}-6x+9 = 0

^{2}+(2m–2)x+(m+2) = 0

D = 0

b

^{2}–4ac = 0

(2m-2)

^{2}–4m(m+2) = 0

4m

^{2}-8m+4–4m

^{2}-8m = 0

–16m = –4

m = ¼

x

^{2}–6x+9 = 0

3. For which „k“ does the equation kx^{2}+(2k+1)x+(k-1)=0 have two different roots in real numbers?

D > 0

b

^{2}–4ac > 0

(2k+1)

^{2}-4k(k-1) > 0

4k

^{2}+4k+1–4k

^{2}+4k > 0

8k > -1

k > –1/8

4. Find „k“ so that the equation x^{2} -5x +k = 0 wouldn‘t have a root in real numbers.

5. Find „m“ so that the equation mx^{2} +2x +m = 0 would have 2 roots in real numbers!

6. Find all „a“, for which one of the equation‘s roots equals null. 2(a-1)x^{2}–(2a-4)x+2a(a-3) = 0. Find out the second root.

7. For which „m“ does the following equation have 1 (double) root in real numbers?

8.

For which „m“ does the equation mx^{2} +(4m –2)x +(4m+1) = 0 have one (double) root in real numbers?

Solution:

The equation has one (double) root in real numbers if and only if m = 1/5.

9.

For which parameters „a“ the equation‘s roots x^{2} – (3a+2)x + a^{2} = 0 fit the relation x_{1} = 9x_{2} ? Determine those roots!

Solution:

10.

For which parameters „a“ in equation 2x^{2} – (a+1)x+(a-1) = 0 fits x_{1} – x_{2} = x_{1} . x_{2} ?

Solution:

The parameter is a = 2, the equation is 2x^{2} – 3x+1 = 0.