x = 17.57 satisfies the given equation
Solution:
Given equation is:
[tex]\sqrt{x^2-10x+25} +12\sqrt{x} =15\sqrt{x}[/tex]
We have to find which values of x will satisfy the given equation
Thus means when we substitute the given value of x and solve the equation, both sides of equation must be same. i,e R.H.S = L.H.S
Substitute x = 1.425 in given equation
[tex]\sqrt{1.425^2-10(1.425)+25} +12\sqrt{1.425} = 15\sqrt{1.425}\\\\\sqrt{2.030625-14.25+25}+14.3248=17.906\\\\\sqrt{12.780625} = 17.906-14.3248\\\\3.575 \neq 3.5812[/tex]
Thus x = 1.425 does not satisfy the given equation
Substitute x = 2.844 in given equation
[tex]\sqrt{2.844^2-10(2.844)+25} +12\sqrt{2.844} = 15\sqrt{2.844}\\\\\sqrt{8.088336-28.44 + 25} +20.236 = 25.296\\\\\sqrt{4.648336} +20.236 = 25.296\\\\2.156 + 20.236 = 25.296\\\\22.392\neq 25.296[/tex]
Thus x = 2.844 does not satisfy the given equation
Substitute x = 16.15 in given equation
[tex]\sqrt{16.15^2-10(16.15)+25} +12\sqrt{16.15} = 15\sqrt{16.15}\\\\\sqrt{260.8225 - 161.5 + 25} + 48.224} = 60.280\\\\\sqrt{124.3225} + 48.224 = 60.280\\\\ 59.374 \neq 60.280[/tex]
Thus x = 16.15 does not satisfies the given equation
Substitute x = 17.57 in given equation
[tex]\sqrt{17.57^2-10(17.57)+25} +12\sqrt{17.57} = 15\sqrt{17.57}\\\\12.57 + 50.2999 = 62.8\\\\62.8 = 62.8[/tex]
Thus x = 17.57 satisfies the given equation
