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Artyom0805 [142]
3 years ago
12

For which of the following equations are x = 5 and x = –5 both solutions?

Mathematics
2 answers:
Minchanka [31]3 years ago
6 0
Short Answer B
Argument
A
A will give you x = +/- 5i 
x^2 + 25 = 0
x^2 = - 25       Take the square root.
sqrt(x^2) = +/- sqrt(-25)
x = +/- (5)i which is a complex number.


B
Is the answer
x^2 = 25
sqrt(x)^2 = sqrt(25)
x = +/- 5

C
Can't be factored just by looking at it. You can show that C is not true just by putting 5 into the equation
f(x) = x^2 + 10x - 25
f(5) = 25 + 10*5 - 25
f(5) = 50
C is not true.

D
D can be eliminated as C was
f(x) = x^2 - 5x - 25
f(5) = -25 ( l'll let you show this is not true). 5 is not a solution because it does not make f(x) = 0
levacccp [35]3 years ago
3 0

x² – 25 = 0 has both x = 5 and x = - 5 as its solutions.

<h3>Further explanation</h3>

Discriminant of quadratic equation ( ax² + bx + c = 0 ) could be calculated by using :

<h2>D = b² - 4 a c</h2>

From the value of Discriminant , we know how many solutions the equation has by condition :

D < 0 → No Real Roots

D = 0 → One Real Root

D > 0 → Two Real Roots

Let us now tackle the problem!

If a quadratic equation has solution x₁ and x₂ , then we could write the equation as following :

\large {\boxed {a (x - x_1)(x - x_2) = 0} }

If x₁ = 5 and x₂ = - 5 , then :

a (x - 5)(x - (-5)) = 0

a (x - 5)(x + 5) = 0

a (x^2 - 5x + 5x - 25) = 0

a (x^2 - 25) = 0

If a = 1 , then we get :

a (x^2 - 25) = 0

1 (x^2 - 25) = 0

\large {\boxed {x^2 - 25 = 0} }

<h3>Learn more</h3>
  • Solving Quadratic Equations by Factoring : brainly.com/question/12182022
  • Determine the Discriminant : brainly.com/question/4600943
  • Formula of Quadratic Equations : brainly.com/question/3776858

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Quadratic Equations

Keywords: Quadratic , Equation , Discriminant , Real , Number

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For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of
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Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is (-1)^{k+1}(\frac{1}{2^{k}}).

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as (-1)^{k+1}(\frac{1}{2^{k}}).

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By the given expression sequence will be \frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......

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and common ratio in each successive term to the previous term is 'r' = \frac{\frac{(-1)}{4}}{\frac{1}{2} }

r = -\frac{1}{2}

Since the sequence is infinite and the formula to calculate the sum is represented by

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S=\frac{\frac{1}{2}}{\frac{3}{2} }

S = \frac{1}{3}

Now we are sure that the sum of infinite terms is \frac{1}{3}.

Therefore, sum of 10 terms will not exceed \frac{1}{3}

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Now we are sure that sum of first 10 terms lie between \frac{1}{4} and \frac{1}{3}

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3 0
3 years ago
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Ann [662]

Answer:

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Step-by-step explanation:

Let the unit place be ‘x'

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Digits are interchanged Unit place' y ‘& tenth place ‘ x'

New no is 10x +y

This is 9 more than the original number

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