Solving an equation via substitution to reduce to quadratic form

The question presents an equation involving radicals, rational expressions, or other non-polynomial terms that requires a substitution (e.g., u = √x, u = x - y) to transform it into a standard quadratic equation to solve.

csat-suneung 2005 Q20 3 marks View
Find the product of all real roots of the irrational equation $x ^ { 2 } + 7 x + 10 + \sqrt { x ^ { 2 } + 7 x + 12 } = 0$. [3 points]
csat-suneung 2009 Q3 3 marks View
What is the sum of all real roots of the equation $\sqrt { x ^ { 2 } - 2 x + 1 } - \sqrt { x ^ { 2 } - 2 x } = \frac { 1 } { 2 }$? [3 points]
(1) 5
(2) 4
(3) 3
(4) 2
(5) 1
csat-suneung 2010 Q19 3 marks View
Find the product of all real roots of the irrational equation $\sqrt { x ^ { 2 } - 7 x + 15 } = x ^ { 2 } - 7 x + 9$. [3 points]
csat-suneung 2011 Q4 3 marks View
For the irrational equation $$\sqrt { 4 x ^ { 2 } - 5 x + 7 } - 4 x ^ { 2 } + 5 x = 1$$ What is the product of all real roots? [3 points]
(1) $- \frac { 1 } { 2 }$
(2) $- \frac { 3 } { 2 }$
(3) $- \frac { 5 } { 2 }$
(4) $- \frac { 7 } { 2 }$
(5) $- \frac { 9 } { 2 }$
csat-suneung 2014 Q24 3 marks View
Find the product of all real roots of the irrational equation $\sqrt { 2 x ^ { 2 } - 6 x } = x ^ { 2 } - 3 x - 4$, and call it $k$. Find the value of $k ^ { 2 }$. [3 points]
csat-suneung 2015 Q24 3 marks View
For the irrational equation $x ^ { 2 } - 6 x - \sqrt { x ^ { 2 } - 6 x - 1 } = 3$, let $k$ be the product of all real roots. Find the value of $k ^ { 2 }$. [3 points]
jee-main 2014 Q61 View
The equation $\sqrt { 3 x ^ { 2 } + x + 5 } = x - 3$, where $x$ is real, has
(1) no solution
(2) exactly four solutions
(3) exactly one solution
(4) exactly two solutions
jee-main 2019 Q61 View
The sum of the solutions of the equation $\sqrt{x} - 2 + \sqrt{x}\sqrt{x} - 4 + 2 = 0, x > 0$ is equal to
(1) 10
(2) 9
(3) 12
(4) 4
jee-main 2023 Q61 View
The number of real solutions of the equation $3 \left( \mathrm { x } ^ { 2 } + \frac { 1 } { \mathrm { x } ^ { 2 } } \right) - 2 \left( \mathrm { x } + \frac { 1 } { \mathrm { x } } \right) + 5 = 0$, is
(1) 4
(2) 0
(3) 3
(4) 2
jee-main 2023 Q64 View
The number of real solutions of the equation $3\left(x^2 + \frac{1}{x^2}\right) - 2\left(x + \frac{1}{x}\right) + 5 = 0$ is
(1) 4
(2) 0
(3) 3
(4) 2
jee-main 2025 Q3 View
The number of solutions of the equation $\left(\frac{9}{x} - \frac{9}{\sqrt{x}} + 2\right)\left(\frac{2}{x} - \frac{7}{\sqrt{x}} + 3\right) = 0$ is:
(1) 2
(2) 3
(3) 1
(4) 4
turkey-yks 2011 Q9 View
$$\frac { 2 ( x - y ) } { x - y - 1 } + \frac { x - y - 1 } { x - y - 2 } = 3$$
Given this, what is the difference $x - y$?
A) $\frac { - 1 } { 2 }$
B) $\frac { - 2 } { 3 }$
C) $\frac { 4 } { 3 }$
D) $\frac { 5 } { 3 }$
E) $\frac { 5 } { 4 }$
turkey-yks 2012 Q9 View
$$x \cdot \left( \sqrt { \frac { 1 } { x } - \frac { 1 } { x ^ { 2 } } } \right) = \frac { 1 } { 2 }$$
Given that, what is x?
A) $\frac { 3 } { 2 }$
B) $\frac { 5 } { 4 }$
C) $\frac { 9 } { 4 }$
D) $\frac { 6 } { 5 }$
E) $\frac { 7 } { 5 }$