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Masteriza [31]
2 years ago
11

What is the root of the quadratic equation y= (x-4) (x+7)

Mathematics
1 answer:
pishuonlain [190]2 years ago
4 0
Root are also known as what x is equal to. So, to find x you would need to set (x-4)=0 and (x+7)=0
X=4. X=-7
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Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0
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Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0

The general equation for the second-order polynomial is:

x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}

This condition must be observed for the case of a quadratic equation with equal roots:

4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0

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k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0

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3 years ago
Please help ASAP! It’s due in 20 min and I have complete it! I just need the statements and reasons! Thank you
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Answer:

1) <11 and <14 are Same-Side Exterior Angles causing them to be supplementary.

2. <8 and <9 are Same-Side Interior Angles causing them to be supplementary.

Step-by-step explanation:

See answer above.

8 0
3 years ago
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