Maths Questions!!!?

Question

1. Show that the equation ax^2 + (a + b) + b = 0 has real roots for all the values of a and b.

2. Find the coordinates of the point that divides the line joining L(-3,-4) to M(3,5) externally in the ratio 3:1

yep i meant ax^2 + (a+b)x + b = 0. thanks!

Answer 1

PROBLEM 1:

I think you meant for (a+b) to be the coefficient on the x-term:
ax² + (a + b)x + b = 0

Real roots are true when the discriminant (item under the sqrt in the quadratic formula) is non-negative.

When you have ax² + bx + c, the discriminant is:
b² - 4ac

In your case your coefficients, in order, are:
a, a+b and b

So the discriminant will be:
(a+b)² - 4(a)(b)

This comes out to:
(a² + 2ab + b²) - 4ab

Simplifying:
a² - 2ab + b²

Then factoring:
(a - b)(a - b)

Or:
(a - b)²

And a square, by definition is always non-negative, which was what you wanted to prove.

PROBLEM 2:

This is similar to finding the midpoint, but you need to instead find the x and y coordinates that are 3/4 of the way between rather than half.

Why 3/4? Because 3:1 means the first segment is 3/4 and the second is 1/4. You put the number over the total.
L(-3,-4) to M(3,5)

Difference of x-coords:
3 - (-3) = 6

Difference of y-coords:
5 - (-4) = 9

3/4 of x distance:
3/4 * 6 = 4½

3/4 of y distance:
3/4 * 9 = 27/4 = 6 3/4

Now you need to add this to the original point L(-3, -4)

The new point P:
x-coord = -3 + 4½ = 1½
y-coord = -4 + 6 3/4 = 2 3/4

P(1 1/2, 2 3/4) divides the line into a ratio of 3 to 1. Draw a picture to convince yourself of this.

Answer 2

2) I forgot the formula. Try it another way.
distance= x1=-3 y1=-4 x2=3 y2=5
=sqrt[(x2-x1)^2+(y2-y1)^2]
=sqrt(6^2+9^2)=sqrt(36+81)=sqrt(117)=10.82
(3/4) of 10.82 is the x coordinate
(1/4) of 10.82 is the y coordinate

1)x=[-(a+b)+-sqrt[(a+b)^2-4ab] /2a
is the discriminant (a+b)^2-4ab >=0
(a+b)^2=a^2+b^2+2ab > 4ab
a^2+b^2 >= 2ab, so it's true

Answer 3

Two questions should be separated. You only get charged 1 point if you give a BEST ANSWER. Don't be cheap when you need homework answers.

PS Thanks for the 2 points!

Answer 4

the two real roots are
y1= -b/a
y2=-1