Use the discriminant to determine how many real number solutions exist for the quadratic equations -4j+3j-28=0
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<h3>Answer: HF = 5</h3>
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Explanation:
Refer to the diagram below. I'm adding in points A and B, which are marked in blue. I'm also adding in variables x, y, and z.
The goal is to find HF, so let's make that equal to x. This is also equal to HB because they are radii of the same circle.
Let y be the length of segments JA and JB.
Let z be the length of segments CF and CA.
So in short we have
- x = HF = HB
- y = JA = JB
- z = CF = CA
We know that
- CH = 13
- HJ = 19
- CJ = 22
This must mean
- CH = CF+HF = z+x = 13
- HJ = HB+JB = x+y = 19
- CJ = CA+JA = z+y = 22
Or in short,
- z+x = 13
- x+y = 19
- z+y = 22
Let's solve the first equation for z to get z = -x+13. I subtracted x from both sides.
Now plug this into the third equation to get
z+y = 22
(z) + y = 22
(-x+13) + y = 22 ... replace z with -x+13
-x+13+y = 22
-x+y+13 = 22
-x+y = 22-13 .... subtract 13 from both sides
-x+y = 9
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So we have this reduced system of equations of two variables and two equations
- x+y = 19 ... found earlier
- -x+y = 9 .... what we just found above
Add the equations straight down. This is the elimination property.
Doing so leads to the x terms canceling out since x + (-x) = 0x = 0
The y terms add to y+y = 2y
The terms on the right hand side add to 19+9 = 28
We are left with 2y = 28 which solves to y = 28/2 = 14.
If y = 14, then,
x+y = 19
x+14 = 19
x = 19-14
x = 5
In which we can then say
z = -x+13
z = -5+13
z = 8
Though we don't need to find z (unless you're curious about it).
Since x = 5, and we set HF equal to x, this means that HF = 5.
