Question 80076: What values of b in the following equations will give one or more real number solutions?
a). 3x^2+bx-3=0
b). 5x^2+bx+1=0
c). -3x^2+bx-3=0
what is a rule for judging if an equation has solutions by looking at it in standard form?
thanks!
Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20055)   (Show Source):
You can put this solution on YOUR website!
What values of b in the following equations will
give one or more real number solutions?
a). 3x^2+bx-3=0
b). 5x^2+bx+1=0
c). -3x^2+bx-3=0
what is a rule for judging if an equation has
solutions by looking at it in standard form?
thanks!
When we have an equation of the form
ax² + bx + c = 0
we form the dicriminant
Discriminant = b² - 4ac
The rule is
There will be at least one real solution
if and only if the discriminant is greater
than or equal 0. There will be one real
solution if the discriminant equals 0 and
2 real solutions if the discriminant is
greater than 0.
a). 3x² + bx - 3 = 0
a = 3, c = -3
b² - 4ac > 0
b² - 4(3)(-3) > 0
b² + 36 > 0
b² > -36
The square of every real number is
always positive or zero and therefore
is always greater than -36
Answer: b can be any real number.
b). 5x² + bx + 1 = 0
a = 5, c = 1
b² - 4ac > 0
b² - 4(5)(1) > 0
b² - 20 > 0
(I will use V for the square root radical)
__ _
critical values are ±V20 = ±2V5 roughly 4.4 and -4.47
------------o----------------o--------------
-2V5 2V5
Substituting a test value below -2V5, b=-5
(-5)² - 20 > 0
25 - 20 > 0
5 > 0
That's true, so we shade the region
of the number line left of -2V5.
<===========o----------------o--------------
-2V5 2V5
Substituting a test value between -2V5 and 2V5, b=0
0² - 20 > 0
0 - 20 > 0
-20 > 0
That's false, so we do not shade the region
of the number line between them.
Substituting a test value above 2V5, b=5
(5)² - 20 > 0
25 - 20 > 0
5 > 0
That's true, so we shade the region
of the number line right of 2V5.
<===========o----------------o=============>
-2V5 2V5
So the values of b for which the solutions are
real are
(-oo,-2V3) U (2V3,oo)
c). -3x^2+bx-3=0
a = -3, c = -3
b² - 4ac > 0
b² - 4(-3)(-3) > 0
b² - 36 > 0
(b - 6)(b + 6) > 0
critical values are 6 and -6
------------o----------------o--------------
-6 6
Substituting a test value below -6, b = -7
(-7)² - 36 > 0
49 - 36 > 0
13 > 0
That's true, so we shade the region
of the number line left of -6.
<===========o----------------o--------------
-6 6
Substituting a test value between -6 and 6, b=0
0² - 36 > 0
0 - 36 > 0
-36 > 0
That's false, so we do not shade the region
of the number line between them.
Substituting a test value above 6, b=7
(7)² - 36 > 0
49 - 36 > 0
13 > 0
That's true, so we shade the region
of the number line right of 6.
<===========o----------------o=============>
-6 6
So the values of b for which the solutions are
real are
(-oo,-6) U (2V3,6)
Edwin
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! What values of b in the following equations will give one or more real number solutions?
Look at b^2-4ac for each equation:
a). 3x^2+bx-3=0
b^2-4ac
=b^2-4*3*-3
=b^2+36
This expression will be non-negative for all values of "b".
-----------
b). 5x^2+bx+1=0
b^2-4ac
=b^2-4*5*1
=b^2-20
This expression will be non-negative for all b>=2sqrt5 or less than -2sqrt5
---------------
c). -3x^2+bx-3=0
b^2-4ac
=b^2-4*-3*-3
=b^2-36
This expression be non-negative for all b >=6 or <= than -6
-------------
what is a rule for judging if an equation has solutions by looking at it in standard form?
b^2>=4ac
===============
Cheers,
Stan H.
|
|
|
