contestada

Which value for the constant d makes x = -2 an extraneous solution in the following equation?
[tex]\sqrt{6-15x} =5x+d[/tex]

Respuesta :

msm555

Answer:

Solution given:

[tex]\sqrt{6-15x} =5x+d[/tex]......(1)

d=?

and

x=-2

Substituting value of x in equation 1.

[tex]\sqrt{6-15*-2} =5*-2+d[/tex].

Solve the like terms

[tex]\sqrt{6+30} =-10+d[/tex].

solve the like terms

[tex]\sqrt{36} =-10+d[/tex].

we have square root of 36 is 6.

6=-10+d

d=16

the value of constant term is 16.

shubhamchouhanvrVT

The value of d for the expression [tex]\sqrt{6-15x} = 5x+d[/tex] is 16.

What is an expression?

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication and division.

The value of 'd' is calculated as,

[tex]\sqrt{6-15x} = 5x+d[/tex]

Put the value of x =-2,

[tex]\sqrt{6-(15\times -2)}=(5\times -2)+d[/tex]

√36 = -10 + d

6 = -10 + d

d = 16

To know more about Expression follow

https://brainly.com/question/723406

#SPJ2