Square root of x plus 8 end root minus 6 equals x

Mathematics

1 answer:

Pachacha [2.7K]2 years ago

5 0

Answer:

Squaring both the sides, we get a quadratic and solving it, we find that the possible values of x are -7 and -4.

Step-by-step explanation:

The expression can be written as √(x+8) - 6 = x. Take 6 on the other side and square both the sides.

[√(x + 8)]² = (x + 6)²

x + 8 = x² + 12x + 36

x² + 11x + 28 = 0

x² + 7x + 4x + 28 = 0

(x + 7)(x + 4) = 0

x = -7 and x = -4

The possible values of x are -7 and -4.

For more explanation, refer the following link:

brainly.com/question/10572241

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You might be interested in

If f(x)=4x2=1 and g(x)=x2-5, find (f-g)(x) A:3x2+6 B:5x2-6 C:3x2-4 D:5x2-4

Alex787 [66]

For this case we have the following functions:

$f (x) = 4x ^ 2 + 1\\g (x) = x ^ 2-5$

We must find $(f-g) (x):$

By definition of operations with functions we have:

$(f-g) (x) = f (x) -g (x)$

So:

$(f-g) (x) = f (x) -g (x) = 4x ^ 2 + 1- (x ^ 2-5) = 4x ^ 2 + 1-x ^ 2 + 5 = 3x ^ 2 + 6$

Answer:

Option A

$3x ^ 2 + 6$

5 0

4 years ago

Number 7. Please asap.

love history [14]

7a)

$\frac{n(n - 3)}{2} = 665\\n(n - 3) = 665 \cdot 2\\n(n - 3) = 1330\\n^2 - 3n = 1330\\n^2 - 3n - 1330 = 0\\n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-1330)}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 + 5320}}{2} = \frac{3 \pm \sqrt{5329}}{2} = \frac{3 \pm 73}{2}\\n_1 = \frac{3 + 73}{2} = \frac{76}{2} = 38\\n_2 = \frac{3 - 73}{2} = \frac{-70}{2} = -35$