Find the Discriminant / Quadratic Formula?

Question

(a) 1= -9x + 11x^2

(b) Base on the answer from part a how many real solution will this equation have??

(c) Solve the following equation using the quadratic formula: 11x^2-9x-1=0

Answer 1

Put it in the form ax^2 + bx + c first so it's clear which terms are which.

a)
11x^2 -9x - 1

Find the discriminate which is b^2 - 4ac
-9^2 - 4*-1*11 = 125

b) Now if the discriminate is
0 you have one real solutions solution
> 0 you have two real solutions
< 0 you have zero real solutions

In this case it's greater than so you have two.

c) just apply the quadratic formula

[-b+/-sqrt(b^2-4ac) ] /2a

[9 +/- sqrt(125)] /22

[9+/- 5sqrt(5) ] / 22

Answer 2

1=-9x+11x^2
11x^2-9x-1=0
The quadratic formula is x={-b+or-rt(b^2-4ac)}/2a,where
a=coefficient of x^2 term
b=coefficient of x term
c=constant

The DISCRIMINANT is the rt(b^2-4ac) term. It is called the discriminant because it can "discriminate" between real solutions and "imaginary" ones, as we shall see in a minute.
For your equation, the discriminant is:
rt{(-9)^2 -4(11)(-1)
rt(81+44)
rt(125)
The number under the sq.rt. sign is positive. You can therefore find the root of a positive number. Since your sq.rt.result must be added to , and subtracted from,b (which in this case is 9), you will get two real, unequal results.
Had the discriminant been equal to 0, you would have got 9+or-0. This looks like only one root, and so it is. In order to keep our sanity, we say that there are 2 real roots, but they are the same. (All quadratic equations have 2 roots)
Had the discriminant been negative, then you can"t go any farther. You cannot take the sq.rt. of a negative number!

For 11x^2-9x-1=0,
a=11
b=-9
c=-1
x={-(-9)+or-rt(-9^2 -4(11)(-1)} /2(11)
x={9+or-rt(81+44)}/22
x={9+or-rt(125)}/22
x={9+or-5rt5}/22
x=(9+5rt5 )/22 or (9-5rt5)/22
You can get "hard" numbers by finding rt5 and doing the arithmetic, but it's OK to leave your answer as I have shown it.
Good luck with your math!

Answer 3

Hi,
Put the equation in standard form:
11x² -9x -1 = 0
Now, based on the equation ax²-bx +c = 0, the discriminant is
b²-4ac
Then the discriminant in you problem is:
= (-9)² -4(11)(-1)
= 81+44
=125

b) This will have two real solutions since, in the quadratic equation the square root term is both added and subtracted from -b:
x= [-b +- sqrt(b²-4ac)]/(2a)

c) Just plug the numbers in the formual I gave you in step b).
x = [-(-9) +- sqrt (125)]/[(2(11)]
=[9 +-5Sqrt(5)]/22
You can do additional arithmetic of your answer requires it.

Hope this helps.
FE

Answer 4

Question a)
11x² - 9x - 1 = 0
x = [ 9 ± √(81 + 44)] / 22
x = [ 9 ± √(125)] / 22
x = [ 9 ± 5√5 ] / 22
Discriminant = 9 ± 5√(5)

Question b)
x = [ 9 ± 5√5] / 22
Two solutions.

Question c)
Solutions are given in part b)

Answer 5

I'm not seeing what you're saying. The discriminant is a way to tell whether the roots are imaginary, real, irrational, etc. It just gives you a heads-up as to what your answer will be like. Then you just substitute it in (So instead of sqrt(b^2 - 4ac), you have the square root of your answer) I hope this information was very helpful.

Answer 6

a) we use (b^2)-4ac
so,
discriminant
=[(-9)^2]-[4(11)(-1)]
=[81]+[44]
=125

b)
since we have discriminant > 0
we have two real roots

c)using -b(+ or -) sq. root of discriminant all over 2a

we get

x = [-(-9)(+-)(sqrt 125)]/2(11)

x = [9(+-)(5sqrt5)]/22

so we have
x=[9+5sqrt5]/22

and

x=[9-5sqrt5]/22