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lozanna [386]
3 years ago
12

All quadratic equations written in standard form can be solved into intercept/factored form prove

Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
3 0

Answer:

Refer below for detailed explanation.

Step-by-step explanation:

1. ax^2+bx+c=0.

Quadratic formula:

2. x = -b+-sqrt(b^2-4ac)/2a. .

Now two equations to find the factored form through these formulas above.

Through this identifying a,b and c.

ax^2+bx+c=0.

x^2+4x-21=0

Here a=1, b=4, and c= -21

Now applying in the quadratic formula.

x=-4+-sqrt(4^2-4(1)(-21))/2(1)

Now solve it and it becomes

x=3 or x= -7.

Another example :

x^2+3x-4=0

Here a=1, b=3 and c=-4.

X=-3+-sqrt(3^2-4(1)(-4))/2(1)

Now solve it and

x= -4 and x=1

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(A)\left(-\dfrac{\sqrt{3} }{2} ,-\dfrac{1 }{2}\right)$ and \left(-\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)\\\\(B)\left(\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)$ and \left(-\dfrac{\sqrt{3} }{2}, \dfrac{1 }{2}\right)\\\\(C)\left(-\dfrac{1 }{2},-\dfrac{\sqrt{3} }{2} \right)$ and \left(\dfrac{1 }{2},\dfrac{\sqrt{3} }{2} \right)\\\\(D)\left(\dfrac{\sqrt{3} }{2},\dfrac{1 }{2} \right)$ and \left(\dfrac{1 }{2},\dfrac{\sqrt{3} }{2} \right)

We observe that only the pair in option C has the same x and y coordinate with the second set of points being a negative factor of the first term. Therefore, they have the same reference angle.

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Step-by-step explanation:

Hope this helps!

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