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2 years ago

Solve the quadratic equation by completing the square. x^2+2x-2=0

Mathematics

1 answer:

astraxan [27]2 years ago

5 0

The solution to the quadratic equation by completing the square is x = -1 + √3, x = -1 - √3

<h3>What is a quadratic equation?</h3>

Any equation of the form $\rm ax^2+bx+c=0$ where x is variable and a, b, and c are any real numbers where a ≠ 0 is called a quadratic equation.

As we know, the formula for the roots of the quadratic equation is given by:

$\rm x = \dfrac{-b \pm\sqrt{b^2-4ac}}{2a}$

We have a quadratic equation:

x² + 2x - 2 = 0

x² + 2x + 1 - 1 - 2 = 0

(x + 1)² - 3 = 0

(x + 1)² = 3

x + 1 = ± √3

x = -1 ± √3

Thus, the solution to the quadratic equation by completing the square is x = -1 + √3, x = -1 - √3

Learn more about quadratic equations here:

brainly.com/question/2263981

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Answer:

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Step-by-step explanation: its right on delta math

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2 years ago

if $5,250 is invested at 4.5% compounded monthly, what is the amount accumulated at the end of 10 years?

Romashka [77]

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In 10 years, you will have $793,789.89

Step-by-step explanation:

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2 years ago

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

2 years ago

Pls help is it A. B. C. D.

Ivenika [448]

Answer:

Step-by-step explanation:

Hope I was able to help!! :)

7 0

3 years ago

X-y=5 and x^2y=5x+6

sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5 (1)

x² · y = 5 · x + 6 (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

To learn more on nonlinear equations: brainly.com/question/20242917

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8 0

1 year ago