Answer:
Inverse of [tex]\bold{y=x^{2}-10 x}[/tex] is [tex]\bold{y=5 \pm \sqrt{x+25}}[/tex]
Solution:
Given that
[tex]y=x^{2}-10 x[/tex]
Adding [tex]5^{2}[/tex] on both sides
[tex]y+5^{2}=x^{2}-10 x+5^{2}[/tex]
Rewrite 10x as 2(5)x,
[tex]y+5^{2}=x^{2}-2(5) x+5^{2}[/tex]
By using [tex]\left(a-b)^{2}=a^{2}-2 a b+b^{2}\right[/tex] , we get
[tex]y+5^{2}=(x-5)^{2}[/tex]
[tex]y+25=(x-5)^{2}[/tex] (Completing the square)
Now swap x and y, we get
[tex]x+25=(y-5)^{2}[/tex]
Rewrite the above equation,
[tex](y-5)^{2}=x+25[/tex]
Taking square root of both sides,
[tex]y-5=\pm \sqrt{x+25}[/tex]
Adding 5 on both sides,
[tex]y=5 \pm \sqrt{x+25}[/tex]
Hence inverse of [tex]\bold{y=x^{2}-10 x \text { is } y=5 \pm \sqrt{x+25}}[/tex]
