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mrs_skeptik [129]
3 years ago
7

What are the solutions to x2 + 8x + 7 = 0

Mathematics
2 answers:
Nezavi [6.7K]3 years ago
8 0
x^{2} +8x+7=0
Product=7
sum=8
x^{2} +x+7x+7=0
x(x+1)+7(x+1)=0
(x+7)(x+1)=0
x=-7, -1
3241004551 [841]3 years ago
6 0
X^2 + 8x + 7 = 0
x^2 + 7x + x + 7 = 0
x(x + 7) + 1(x + 7) = 0
(x + 7) (x + 1) = 0

x = -7      or       x = -1
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Solve the following equation for x. x = -2 x = 2 x = -17 x = -7 (x - 5)/2 = -6
nordsb [41]

Step-by-step explanation:

x= -2,

(x - 5) / 2= -6

( (-2) - 5) / 2= -6

(-7) / 2 = -6

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right hand side not equal to left hand side of this equation.so,x= -2 cannot exist for this equation.

x=2,

(x - 5) / 2= -6

(2 - 5) / 2= -6

(-3) / 2= -6

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right hand side not equal to left hand side of this equation.so,x= 2 cannot exist for this equation.

x= -17

(x - 5) / 2= -6

( ( -17) - 5) / 2= -6

(- 22) / 2= -6

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right hand side not equal to left hand side of this equation.so,x= -17 cannot exist for this equation.

x= -7,

(x - 5) / 2= -6

( ( -7) -5) / 2= -6

(-12) / 2= -6

-6= -6

right hand side equal to left hand side of this equation.so,x= -7 exist for this equation.

8 0
2 years ago
Find the third side in simplest radical form:<br> 25
Gre4nikov [31]

Answer:

<h3>\boxed{  \bold{24}}</h3>

Step-by-step explanation:

\mathsf{given}

\mathsf{hypotenuse(h) = 25}

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\sf{base(b) = }?

Now, Using Pythagoras theorem

\sf{{h}^{2}  =  {p}^{2}  +  {b}^{2} }

plug the values

⇒\sf{  {25}^{2}  =  {7}^{2}  +  {b}^{2} }

Evaluate the power

⇒\sf{625 = 49 +  {b}^{2} }

Swap the sides of the equation

⇒\sf{49 +  {b}^{2}  = 625}

Move constant to right hand side and change it's sign

⇒\sf{ {b}^{2}  = 625 - 49}

Calculate the difference

⇒\sf{ {b}^{2}  = 576}

Squaring on both sides

⇒\sf{b = 24}

Hope I helped!

Best regards!

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Answer:

<h2>2/5 </h2>

Step-by-step explanation:

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