Answer:
Option B
Step-by-step explanation:
We have to solve this system of equations by using the Substitution method so here goes,
[tex]Equation\ 1\\\\\y=x^2+5x+3\\\\Equation\ 2\\\\\ y=\sqrt{2x+5}\\[/tex]
We insert equation 2 in equation 1 since both equations are equal to y so we equate both of them,
[tex]\sqrt{2x+5}=x^2+5x+3\\Square\ both\ sides\\\\\2x+5=(x^2+5x+3)^2 \\[/tex]
To square the right hand side suppose
[tex]a=x^2+5x\\b=3\\(a+b)^2=a^2+2ab+b^2\\(x^2+5x)^2+2(x^2+5x)(3)+(3)^2\\\\((x^2)^2+2(x^2)(5x)+(5x)^2)+6(x^2+5x)+9\\x^4+10x^3+25x^2+6x^2+30x+9\\x^4+10x^3+31x^2+30x+9\\[/tex]
Now,
[tex]2x+5=x^4+10x^3+31x^2+30x+9\\x^4+10x^3+31x^2+30x-2x+9-5=0\\x^4+10x^3+31x^2+28x+4=0\\[/tex]
Apply Newton-Raphson method to quickly determine the approximate solution of the polynomial we get the values of x to be,
x ≈ -0.175 and x ≈ -1.22
now insert these values into equation 1 or 2 to get the values of y
we use the first value ,
[tex]y=x^2+5x+3\\y=(-0.175)^2+5(-0.175)+3\\y=2.155\\[/tex]
now for the second value of x,
[tex]y=x^2+5x+3\\y=(-1.22)^2+5(-1.22)+3\\y=-1.612[/tex]
now since the second set of values are not in the options we ignore the second set of values which are ( -1.22 , -1.612)
we use the first set of values which are (-0.175 , 2.155) which rounded off lead to (-0.2 , 2.1), so Option B is your answer
I also attached a graphical representation so you can verify the answer yourself , The approximate solution of this system of equation is the point of intersection between both the graphs which is marked on the image.
