Help me ! Geometry and Algebra!

bingbong · 2016-03-04 18:10:27

Hey, I would like some help on this geometry problem, the full answer is fine as long as theres an explanation too!
Two lines l and m intersect at O at an angle of 28^\circ. Let A be a point inside the acute angle formed by l and m. Let B and C be the reflections of A in lines l and m, respectively. Find the number of degrees in \angle BAC.

Also, I am stuck on this algebra problem:
p and q are the solutions to the quadratic equation x^2 + 4x + 6 = 0.

p^2 and q^2 are the solutions to the quadratic equation x^2 + bx + c = 0.

Find b + c.

Thanks in advance!

Last edited by bingbong (2016-03-04 18:11:38)

Nehushtan · 2016-03-04 19:13:21

Hints:

(1) Angle BOC = 2×28° (why?). Calculate the sum of the angles ABO and ACO. (NB: You don't need to calculate them individually, only their sum.) These will enable you to work out angle BAC in the quadrilateral BOAC.

(2)

Last edited by Nehushtan (2016-03-04 19:18:01)

Grantingriver · 2016-03-05 06:07:06

1) Alternative and simpler solution: since B and C are the reflections of A in the lines l and m respectively, then the lines AB and AC form right angles with l and m respectively. Now if we place the point A any where inside the angle 28˚ the angle BÂC will not be altered, since it will lie on the intersection of the lines AB and AC or on the intersection of two lines parallel to them. So to simplify the problem, let's place it in the line which bisect the angle 28˚ and hence bisect the angle BÂC, therefore we will have two identical trangles, in both of them, two angles are known and the thrid angle is half the angle BÂC, therefore since the sum of the angles of a plane trangle is 180˚ then:
0.5×BÂC=180˚-(90˚ +0.5×28˚) ⇒ BÂC=2×76=152˚

2) Alternative and obvious solution: since p² and q² are the solutions of x²+bx+c=0 then:
(p²)²+b×p²+c=(q²)²+b×q²+c ⇒ (p⁴-q⁴)=-b(p²-q²) ⇒ -b=(p²-q²)(p²+q²)/(p²-q²)=(p²+q²)
Now substituting b=-(p²+q²) in the original equation we have:
(p²)²-(p²+q²)×p²+c=0 ⇒ p⁴-p⁴-p²q²+c=0 ⇒ c= p²q²
Now after solving the other equation (which is easy by using the general law) you can find the required quantity which is "b+c=p²q²-(p²+q²)".

bingbong · 2016-03-05 09:00:52

Thanks guys, I got the solution to the problems!

Nehushtan · 2016-03-05 10:14:18

Grantingriver wrote:

2) Alternative and obvious solution: since p² and q² are the solutions of x²+bx+c=0 then:
(p²)²+b×p²+c=(q²)²+b×q²+c ⇒ (p⁴-q⁴)=-b(p²-q²) ⇒ -b=(p²-q²)(p²+q²)/(p²-q²)=(p²+q²)
Now substituting b=-(p²+q²) in the original equation we have:
(p²)²-(p²+q²)×p²+c=0 ⇒ p⁴-p⁴-p²q²+c=0 ⇒ c= p²q²
Now after solving the other equation (which is easy by using the general law) you can find the required quantity which is "b+c=p²q²-(p²+q²)".

The answer is a numerical value, not in terms of p and q.

Math Is Fun Forum

#1 2016-03-04 18:10:27

Help me ! Geometry and Algebra!

#2 2016-03-04 19:13:21

Re: Help me ! Geometry and Algebra!

#3 2016-03-05 06:07:06

Re: Help me ! Geometry and Algebra!

#4 2016-03-05 09:00:52

Re: Help me ! Geometry and Algebra!

#5 2016-03-05 10:14:18

Re: Help me ! Geometry and Algebra!

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