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dem82 [27]
3 years ago
6

Explain why a positive discriminant results in two real solutions...

Mathematics
1 answer:
marta [7]3 years ago
3 0
You have three cases for a discriminant:

1. Zero 
A zero in the discriminant means that there is one solution, since it's ... plus/minus zero, which is just ... 

2. Negative
Negative means no real solutions, as the square root of a negative number is not real. 

3. Positive
If you have a positive number within the square root, it means that your solutions will be ... plus/minus whatever the square root is. One solution is the plus, and one solution is the minus, hence two solutions. 

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Find the intersection point between the lines of equations:<br><br>2x-y+6=0 and 2x+3y-6=0 ​
Ugo [173]

Step-by-step explanation:

The two equation will intersect each other at the point which will be the solution of the given two equations , and the given equations are ,

\implies 2x -y +6=0\\\\\implies 2x + 3y -6=0

On subtracting the given equations we have,

\implies -y - 3y +6 -(-6) = 0 \\\\\implies -4y = -12 \\\\\implies y = -12/-4\\\\\implies y = 3

Put this value in any equation , we have ,

\implies 2x -3 +6 =0\\\\\implies 2x = -3 \\\\\implies x =\dfrac{-3}{2} \\\\\implies x =-1.5

Hence the lines will Intersect at ,

\implies\underline{\underline{ Point=(-1.5, 3)}}

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9.6 or 9 and 3/5

Step-by-step explanation:

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Solving Rational equations. LCD method. Show work. Image attached.
posledela

(k-2)(k-6)=k^2-2k-6k+12=k^2-8k+12

So in order to get all the fractions to have a common denominator, we need to multiply \dfrac k{k-2} by \dfrac{k-6}{k-6}, and \dfrac1{k-6} by \dfrac{k-2}{k-2}:

\dfrac k{k-2}\cdot\dfrac{k-6}{k-6}=\dfrac{k(k-6)}{(k-2)(k-6)}=\dfrac{k^2-6k}{k^2-8k+12}

\dfrac1{k-6}\cdot\dfrac{k-2}{k-2}=\dfrac{k-2}{(k-2)(k-6)}=\dfrac{k-2}{k^2-8k+12}

Now,

\dfrac4{k^2-8k+12}=\dfrac{(k^2-6k)+(k-2)}{k^2-8k+12}

As long as k\neq2 and k\neq6 (which we can't have because otherwise k^2-8k+12=0), we can cancel k^2-8k+12 in the denominators on both sides:

4=(k^2-6k)+(k-2)

4=k^2-5k-2

0=k^2-5k-6

We can factorize the right side:

0=(k-6)(k+1)

which tells us that k=6 and k=-1 are solutions.

4 0
4 years ago
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