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expeople1 [14]
2 years ago
15

Which are the solutions of the quadratic equation? x2 = –5x – 3

Mathematics
2 answers:
Oksanka [162]2 years ago
5 0
I think the answer is 0.541 I am not sure 
quester [9]2 years ago
4 0
The goal of solving a quadratic equation:

x^{2} = -5x - 3

Move all terms to one side and set the equation = to zero. This means adding -5x and 3 to both sides.

x^{2} + 5x + 3 = 0

To solve a quadratic equation, you need the quadratic formula. The purpose of the quadratic formula is to find the "zeros" of an equation when the "zeros" are not whole numbers. A "zero" of an equation is the value of x when y equals zero, so basically the x-intercept. The x-intercepts are the points where your graph will intersect the x-axis.

Quadratic formula: \frac{-b\frac{+}{-} \sqrt{(b)^{2} - 4ac}}{2a}

\frac{+}{-} means "plus" or "minus." if there are two signs, there will be two answers.

In the quadratic formula, you may have observed the values of a, b, and c.

These values can be found in the standard form of an exponential graph.

Standard Form: ax^{2} +bx + c

"a" is the coefficient of the first term in your original equation.

x^{2} + 5x + 3 = 0

The coefficient of x^{2} is 1.

"b" is the coefficent of the second term.

The coefficient of 5x is 5.

"c" is your constant term, which is a value that does not change. In this case, it is 3, the number at the end.

Plug in your a, b, and c values into the quadratic equation.

a = 1, b = 5, c = 3

\frac{-(5)\frac{+}{-} \sqrt{(5)^{2} - 4(1)(3)}}{2(1)}

\frac{-5\frac{+}{-} \sqrt{(25 - 12)}}{2}

\frac{-5\frac{+}{-} \sqrt{(13)}}{2}

This expression can be considered the final answer. However, in some cases, you will need to simplify further. Because of the plus/minus symbol, you will have two distinct answers.

\frac{-5\ - \sqrt{13}}{2}     \frac{-5\ + \sqrt{13}}{2}

One answer has a "-" symbol and the other answer has a "+" symbol. This is because a square root will always have two answers: a negative number and a positive number. \sqrt{x} times \sqrt{x} = x. -\sqrt{x} times -\sqrt{x} = x.

The square root of 13 is roughly 3.6

\frac{-5 - 3.6}{2}     \frac{-5 + 3.6}{2}

\frac{-8.6}{2}           \frac{-1.4}{2}

x = -4.3, x = -0.7

I know my answer is long, but the process is very straight forward. Simply find your a, b, and c values, plug them into the quadratic formula, and simplify.

Some common mistakes made in this process include:

Not including the negative in you a, b, and c values. For example:

x^{2} -3x - 10

a = 1, b = -3, and c = -10.

Remember to include the negative sign! It is part of the value itself.

Another mistake is not setting the equation to zero. Perhaps there is a number or term on the other side of the equation. For example:

x^{2} -3x - 5 = 5

Do NOT leave the 5 on the other side. The equation must be equal to zero. Remember we are trying to find the x-intercepts? To find those, y must = 0.

Subtract or add to get rid of the terms on the other side.


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sesenic [268]

Answer:

4 \cot {}^{2} (45)  - 3 \tan {}^{2} (60)  + 2 \sec {}^{2} (60)  = 3 \\

LHS = 4 \cot {}^{2} (45)  - 3 \tan {}^{2} (60)  + 2 \sec {}^{2} (60)  \\

let us first take a look at the values of the trigonometric ratios given in the question so that we get quite clear about what is to be done.

here ,

\cot(45)  = 1 \\  \\  \tan(60)  =  \sqrt{3}  \\  \\  \sec(60)  = 2

now ,

we just have to plug in the values considering certain other things given in the question and we're done!

so let's start ~

4(1) {}^{2}  - 3( \sqrt{3} ) {}^{2}  + 2(2) {}^{2}  \\ \\ \dashrightarrow \: 4(1) - 3(3) + 2(4) \\ \\ \dashrightarrow \: 4 - 9 + 8 \\\\ \dashrightarrow \: 12 - 9 \\ \\\dashrightarrow \: 3 = RHS

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hope helpful :D

7 0
2 years ago
What is the remainder of <br> (x3 – 6x – 9x + 3) ÷ (x - 3)
daser333 [38]
<span> remainder is f(3) </span>
<span>3^3 - 6*3^2 - 9*3 + 3 </span>
<span>= 27 - 54 - 27 + 3 </span>
<span>= -51</span>
6 0
3 years ago
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Determine if the lim f(x) exists using the graph below.. If it does, find its value. If it does not,x 1explain why
Leno4ka [110]

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Also,

\begin{gathered} \text{ From the given image, the value of the right limit of f at x=1 } \\ \lim_{x\to a^+}f(x)\gt1 \end{gathered}

Therefore,

\lim_{x\to1^-}f(x)\lim_{x\to1^+}f(x)

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8 0
10 months ago
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