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Elza [17]
3 years ago
6

Time left: 0:48:24Question 11 of 50What are the roots of the quadratic equationx2+3x+2?​

Mathematics
2 answers:
TiliK225 [7]3 years ago
8 0

\boxed{\huge{ \mathfrak{  { \:  \:    \: \:  \:  \: Answer \:  \:   \:    \:  \: }   } }}

Let's Solve :

  • {x}^{2}  + 3x + 2

  • {x}^{2}  + 2x + x + 2

  • x(x + 2) + 1(x + 2)

  • (x + 2)(x + 1)

Now let's Equate it with 0

Case - 1 : where, x + 1 = 0

  • x = -1

Case - 2 : where, x + 2 = 0

  • x = -2

Therefore the roots of the given quadratic equation is -1 and -2

_____________________________

\mathrm{ \#TeeNForeveR}

alexgriva [62]3 years ago
6 0

Answer:

-1 & -2

Step-by-step explanation:

x²+3x+2

x²+2x+x+2

(x²+2x)+(x+2)

x(x+2)+1(x+2)

(x+1)(x+2)

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Answer:

16

Step-by-step explanation:

The distance is given by

d = sqrt( (x2-x1) ^2 + ( y2-y1) ^2)

Since the x values are the same

d = sqrt(  ( y2-y1) ^2)

d = sqrt(  (( -6 - 10)^2))

d =  sqrt((-16) ^2)

d = 16

We can also just find the number of squares between the points since the x value is the same.

10 - -6

10+6

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I need help 6x-2y=10 x-2y=-5 solve by elimination
dem82 [27]
<h3><u>Explanation</u></h3>
  • Given the system of equations.

\begin{cases} 6x - 2y = 10 \\ x - 2y =  - 5 \end{cases}

  • Solve the system of equations by eliminating either x-term or y-term. We will eliminate the y-term as it is faster to solve the equation.

To eliminate the y-term, we have to multiply the negative in either the first or second equation so we can get rid of the y-term. I will multiply negative in the second equation.

\begin{cases} 6x - 2y = 10 \\  - x  +  2y =  5 \end{cases}

There as we can get rid of the y-term by adding both equations.

(6x - x) + ( - 2y + 2y) = 10 + 5 \\ 5x + 0 = 15 \\ 5x = 15 \\ x =  \frac{15}{5}  \longrightarrow  \frac{ \cancel{15}}{ \cancel{5}}  =  \frac{3}{1}  \\ x = 3

Hence, the value of x is 3. But we are not finished yet because we need to find the value of y as well. Therefore, we substitute the value of x in any given equations. I will substitute the value of x in the second equation.

x - 2y =  - 5 \\ 3 - 2y =  - 5 \\ 3 + 5 = 2y \\ 8 = 2y \\  \frac{8}{2}  = y \\ y =  \frac{8}{2} \longrightarrow  \frac{ \cancel{8}}{ \cancel{2}}  =  \frac{4}{1}  \\ y = 4

Hence, the value of y is 4. Therefore, we can say that when x = 3, y = 4.

  • Answer Check by substituting both x and y values in both equations.

<u>First</u><u> </u><u>Equation</u>

6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10  = 10 \longrightarrow \sf{true} \:  \green{ \checkmark}

<u>Second</u><u> </u><u>Equation</u>

x - 2y =  - 5 \\ 3 - 2(4) =  - 5 \\ 3 - 8 =  - 5 \\  - 5 =  - 5 \longrightarrow  \sf{true} \:  \green{ \checkmark}

Hence, both equations are true for x = 3 and y = 4. Therefore, the solution is (3,4)

<h3><u>Answer</u></h3>

\begin{cases} x = 3 \\ y = 4 \end{cases} \\  \sf \underline{Coordinate \:  \: Form} \\ (3,4)

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Shalnov [3]

i think the answer is d

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