Question

The equation $$x - \frac{2}{{x - 1}} = 1 - \frac{2}{{x - 1}}\,$$     has

A. no root  
B. one root
C. two equal roots
D. infinitely many roots
Answer :   no root
Solution :
Given equation is $$x - \frac{2}{{x - 1}} = 1 - \frac{2}{{x - 1}}$$
Clearly $$x \ne 1$$  for the given eqn. to be defined. If $$x - 1 \ne 0,$$   we can cancel the common term $$\frac{{ - 2}}{{x - 1}}$$ on both sides to get $$x = 1,$$  but it is not possible. So given eqn. has no roots.
∴ $${\text{(A)}}$$ is the correct answer.

Releted MCQ Question on
Algebra >> Quadratic Equation

Releted Question 1

If $$\ell ,m,n$$  are real, $$\ell \ne m,$$  then the roots by the equation: $$\left( {\ell - m} \right){x^2} - 5\left( {\ell + m} \right)x - 2\left( {\ell - m} \right) = 0$$         are

A. Real and equal
B. Complex
C. Real and unequal
D. None of these
Releted Question 2

The equation $$x + 2y + 2z = 1{\text{ and }}2x + 4y + 4z = 9{\text{ have}}$$

A. Only one solution
B. Only two solutions
C. Infinite number of solutions
D. None of these
Releted Question 3

Let $$a > 0, b > 0$$    and $$c > 0$$ . Then the roots of the equation $$a{x^2} + bx + c = 0$$

A. are real and negative
B. have negative real parts
C. both (A) and (B)
D. none of these
Releted Question 4

Both the roots of the equation $$\left( {x - b} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - c} \right) + \left( {x - a} \right)\left( {x - b} \right) = 0$$           are always

A. positive
B. real
C. negative
D. none of these.

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Quadratic Equation


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