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Determine all the real roots of the equation...

(x+7)(x^2- 49)=0

Drag and drop all real zeros of the equation into the box.

(These were the options that I were given.)

-49, -14, -7, -4, -3, 0, 3, 4, 7, 14, 49

Respuesta :

Let's solve it!

So, let's study each factor.

x + 7 = 0 -> x = -7

x² - 49 = 0 -> x² = 49 - > x = ± √49 -> x = ± 7

Final answer: x = 7 and x = -7

The real roots of the equation are -7 and 7. Among the given options, only -7 and 7 are correct.

What are the real roots of an equation?

Real roots of an equation are real numbers that satisfy the equation.

From the question, we are to determine the roots of the given equation.

The equation is
(x+7)(x²- 49)=0

This means

x+7 = 0 OR x²- 49 = 0

∴ x = -7 OR (x+7)(x-7) = 0

NOTE: By applying difference of two squares, x²- 49 = (x+7)(x-7)

Then,

x = -7 OR x+7 = 0 OR x-7 = 0

x = -7 OR x = -7 OR x = 7

Thus,

x = -7 OR 7

Hence, the real roots of the equation are -7 and 7. Among the given options, only -7 and 7 are correct.

Learn more on Determining the roots of an equation here: https://brainly.com/question/2193153

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