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2 years ago

X^2+6x-16=0 quadratic equation

Mathematics

2 answers:

Black_prince [1.1K]2 years ago

4 0

（x+2）（x-8）is the answer

Aleonysh [2.5K]2 years ago

3 0

Answer:

x = - 8, x = 2 are the solutions

Step-by-step explanation:

Given

x² + 6x - 16 = 0

Consider the factors of the constant term (- 16) which sum to give the coefficient of the x- term (+ 6)

The factors are + 8 and - 2 , since

8 × - 2 = - 16 and 8 - 2 = + 6 , thus

(x + 8)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 8 = 0 ⇒ x = - 8

x - 2 = 0 ⇒ x = 2

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makkiz [27]

Answer:

The series converges to $\frac{1}{1-9x}$ for $\frac{-1}{9} < x < \frac{1}{9}$

Step-by-step explanation:

Given the series is $\sum_{n=0}^{\infty} 9^n x^n$

We have to find the values of x for which the series converges.

We know,

$\sum_{n=0}^{\infty} ar^{n-1}$ converges to (a) / (1-r) if r < 1

Otherwise the series will diverge.

Here, $\sum_{n=0}^{\infty} 9^n x^n = \sum_{n=0}^{\infty} (9x)^{n}$ is a geometric series with |r| = | 9x |

And it converges for |9x| < 1

Hence, the given series gets converge for $\frac{-1}{9} < x < \frac{1}{9}$

And geometric series converges to $\frac{a}{1-r}$

Here, a = 1 and r = 9x

Therefore, $\frac{a}{1-r} = \frac{1}{1-9x}$

Hence, the given series converges to $\frac{1}{1-9x}$ for $\frac{-1}{9} < x < \frac{1}{9}$

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3 years ago