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MrMuchimi
3 years ago
8

Solving Quadratic Equations using the Square Root Property

Mathematics
1 answer:
Afina-wow [57]3 years ago
5 0

Answer:

x = -2   or   x = -5

Step-by-step explanation:

You need to complete the square before you can take the square root of both sides.

x^2 + 7x + 10 = 0

Subtract 10 from both sides.

x^2 + 7x = -10

To complete the square, you need to add the square of half of the x-term coefficient to both sides.

The x-term coefficient is 7. Half of that is 7/2. Square it to get 49/4. Now we add 49/4 to both sides of the equation.

x^2 + 7x + \dfrac{49}{4} = -10 + \dfrac{49}{4}

(x + \dfrac{7}{2})^2 = -\dfrac{40}{4} + \dfrac{49}{4}

(x + \dfrac{7}{2})^2 = \dfrac{9}{4}

Now we use the square root property, if

x^2 = k, then

x = \pm \sqrt{k}

x + \dfrac{7}{2} = \pm \sqrt{\dfrac{9}{4}}

x + \dfrac{7}{2} = \pm \dfrac{3}{2}

x + \dfrac{7}{2} = \dfrac{3}{2}   or   x + \dfrac{7}{2} = -\dfrac{3}{2}

x = -\dfrac{4}{2}   or   x = -\dfrac{10}{2}

x = -2   or   x = -5

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The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)

The coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

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We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.

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