S is the midpoint of RT RS =x2+3 ST =−6x+10.
The possible values of RT are 8 , 104
Given :
S is the midpoint of RT RS =x2+3 ST =−6x+10
S is the midpoint , so the length of RS and ST are same .
Make the expression equal to each other and solve for x
[tex]RS=ST\\x^2+3=-6x+10\\x^2+3-10=-6x+10-10\\x^2-7=-6x\\x^2-7+6x=-6x+6x\\x^2+6x-7=0\\[/tex]
Solve for x using quadratic formula
[tex]x=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \left(-7\right)}}{2\cdot \:1}\\x=\frac{-6\pm \:8}{2\cdot \:1}\\x=\frac{-6+8}{2\cdot \:1},\:x=\frac{-6-8}{2\cdot \:1}\\x=1, -7[/tex]
Now we find out RT when x=1 and x=-7
[tex]RT=RS+ST\\RT=x^2+3-6x+10\\RT=x^2-6x+13\\\\x=1\\RT=1^2-6(1)+13=8\\\\x=-7\\RT=\left(-7\right)^2-6\left(-7\right)+13=104[/tex]
The possible values of RT are 8 , 104
Learn more : brainly.com/question/10895180
