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katrin [286]
2 years ago
9

Solve for x in the equation: x^2 - x = 20

Mathematics
1 answer:
max2010maxim [7]2 years ago
8 0
X=5
cuz 5^2=25
25-5=20
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What are the solutions of x^2-2x+5=0
sweet [91]

The solution of x^{2}-2 x+5=0 are 1 + 2i and 1 – 2i

<u>Solution:</u>

Given, equation is x^{2}-2 x+5=0

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}  --- (1)

<em><u>Let us determine the nature of roots:</u></em>

Here in x^{2}-2 x+5=0 a = 1 ; b = -2 ; c = 5

b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16

Since b^2 - 4ac < 0 , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}

On solving we get,

x=\frac{2 \pm \sqrt{4-20}}{2}

x=\frac{2 \pm \sqrt{-16}}{2}

x=\frac{2 \pm \sqrt{16} \times \sqrt{-1}}{2}

we know that square root of -1 is "i" which is a complex number

\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}

Hence, the roots of the given quadratic equation are 1 + 2i and 1 – 2i

6 0
2 years ago
HELP ME NOW!!!!!!!!!! (ALGEBRA)
melisa1 [442]

\\ \tt\hookrightarrow( g\circ h)(x)=g(h(x))

\\ \tt\hookrightarrow \dfrac{8(3x+10)-1}{3x+10+4}

\\ \tt\hookrightarrow \dfrac{8(3x)+8(10)-1}{3x+14}

\\ \tt\hookrightarrow \dfrac{24x+80-1}{3x+14}

\\ \tt\hookrightarrow \dfrac{24x+79}{3x+14}

Done

3 0
2 years ago
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Nces
saul85 [17]

Answer:

d. am

Step-by-step explanation:

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8 0
2 years ago
20 grams of KCl to moles of
Irina-Kira [14]
0.013413582325191 got it from google
4 0
3 years ago
Two numbers multiply to be -7 and add to be -6. What are the numbers?<br><br> Example; 2,3
rewona [7]

<u>Given</u>:

Let the two numbers be x and y.

Two numbers multiply to be -7 and add to be -6.

This can be written in equation as,

xy=-7 and

x+y=-6

<u>Value of the two numbers:</u>

Let us determine the value of the two numbers using substitution method.

Substituting y=-6-x in the equation xy=-7, we get;

x(-6-x)=-7

Simplifying, we get;

       -6x-x^2=-7

   x^2+6x-7=0

(x+7)(x-1)=0

x=-7, 1

Thus, the values of x are x = 1,-7

When x = 1 , the equation x+y=-6 becomes y=-7

When x = -7, the equation x+y=-6 becomes y=1

Therefore, the two numbers are 1 and -7

3 0
3 years ago
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