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4 years ago

Which equation has roots of 3plus or minus sqare rt 2?

Mathematics

1 answer:

vichka [17]4 years ago

6 0

Answer:

Option D is correct.

The equation with roots 3 plus or minus square root 2 is x² - 6x + 7

Step-by-step explanation:

The roots of the unknown equation are

3 ± √2, that is, (3 + √2) and (3 - √2)

The equation can then be reconstructed by writing these roots as the solutions of the quadratic equation

x = (3 + √2) or x = (3 - √2)

The equation is this

[x - (3 + √2)] × [x - (3 - √2)]

(x - 3 - √2) × (x - 3 + √2)

x(x - 3 + √2) - 3(x - 3 + √2) - √2(x - 3 + √2)

= x² - 3x + x√2 - 3x + 9 - 3√2 - x√2 + 3√2 - 2

Collecting like terms

= x² - 3x - 3x + x√2 - x√2 - 3√2 + 3√2 + 9 - 2

= x² - 6x + 7

Hope this Helps!!!

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The perpendicular bisector of the line segment connecting the points (-3,8) and (-5,4) has an equation of the form y = mx + b. F

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Answer:

Step-by-step explanation:

find the slope

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this means that the equation is y=2x-13

and if you add m and b

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I HOPE THIS HELPS

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zavuch27 [327]

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4 years ago

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Find the two intersection points

bogdanovich [222]

Answer:

Our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

Step-by-step explanation:

We want to find where the two graphs given by the equations:

$\displaystyle (x+1)^2+(y+2)^2 = 16\text{ and } 3x+4y=1$

Intersect.

When they intersect, their x- and y-values are equivalent. So, we can solve one equation for y and substitute it into the other and solve for x.

Since the linear equation is easier to solve, solve it for y:

$\displaystyle y = -\frac{3}{4} x + \frac{1}{4}$

Substitute this into the first equation:

$\displaystyle (x+1)^2 + \left(\left(-\frac{3}{4}x + \frac{1}{4}\right) +2\right)^2 = 16$

Simplify:

$\displaystyle (x+1)^2 + \left(-\frac{3}{4} x + \frac{9}{4}\right)^2 = 16$

Square. We can use the perfect square trinomial pattern:

$\displaystyle \underbrace{(x^2 + 2x+1)}_{(a+b)^2=a^2+2ab+b^2} + \underbrace{\left(\frac{9}{16}x^2-\frac{27}{8}x+\frac{81}{16}\right)}_{(a+b)^2=a^2+2ab+b^2} = 16$

Multiply both sides by 16:

$(16x^2+32x+16)+(9x^2-54x+81) = 256$

Combine like terms:

$25x^2+-22x+97=256$

Isolate the equation:

$\displaystyle 25x^2 - 22x -159=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 25, b = -22, and c = -159. Substitute:

$\displaystyle x = \frac{-(-22)\pm\sqrt{(-22)^2-4(25)(-159)}}{2(25)}$

Evaluate:

$\displaystyle \begin{aligned} x &= \frac{22\pm\sqrt{16384}}{50} \\ \\ &= \frac{22\pm 128}{50}\\ \\ &=\frac{11\pm 64}{25}\end{aligned}$

Hence, our two solutions are:

$\displaystyle x_1 = \frac{11+64}{25} = 3\text{ and } x_2 = \frac{11-64}{25} =-\frac{53}{25}$

We have our two x-coordinates.

To find the y-coordinates, we can simply substitute it into the linear equation and evaluate. Thus:

$\displaystyle y_1 = -\frac{3}{4}(3)+\frac{1}{4} = -2$

And:

$\displaystyle y _2 = -\frac{3}{4}\left(-\frac{53}{25}\right) +\frac{1}{4} = \frac{46}{25}$

Thus, our two intersection points are:

$\displaystyle (3, -2) \text{ and } \left(-\frac{53}{25}, \frac{46}{25}\right)$

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Answer:

Step-by-step explanation:

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