$f(x) =x^2 -2x\:\: and \:\: g(x) = 6x + 4\\(f+g)(x) = f(x) + g(x) \\(f+g)(x) = x^2 -2x+ 6x + 4\\(f+g)(x) = x^2+4x + 4\\(f+g)(x) = (x+2)^2\\ 0 = (x+2)^2...[\because (f+g)(x) =0]\\0 = x+2\\\therefore x = -2$

6 0

3 years ago

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Find a cubic function with the given zeros. (1 point) Square root of seven. , - Square root of seven. , -4

Debora [2.8K]

Given :

Three roots , $\sqrt{7},\ -\sqrt{7},\ -4$ .

To Find :

A cubic function with the given zeros.

Solution :

Equation of polynomial of 3 zeroes is given by :

$(x-a_1)(x-a_2)(x-a_3)=0\\\\(x-\sqrt{7})(x+\sqrt{7})(x+4)=0\\\\( x^2-7)(x+4)=0\\\\x^3+4x^2-7x-28=0$

Therefore, the cubic function of given zeroes is $x^3+4x^2-7x-28=0$ .

Hence, this is the required solution.

6 0

4 years ago

Need help as fast as possible please

ankoles [38]

Answer:

Correct Answer

Step-by-step explanation:

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4 0

3 years ago