the line y 14 0 cuts the curve whose equation is x x 2 x 1

The line $$y + 14 = 0$$ cuts the curve whose equation is $$x\left( {{x^2} + x + 1} \right) + y = 0$$ at

A. three real points

B. one real point

C. at least one real point

D. no real point

Answer : one real point

Solution :
Solving the equations, $${x^3} + {x^2} + x - 14 = 0$$
$$\eqalign{ & \Rightarrow \,\,\left( {x - 2} \right)\left( {{x^2} + 3x + 7} \right) = 0 \cr & \therefore \,\,x = 2,\frac{{ - 3 \pm \sqrt {9 - 28} }}{2}. \cr} $$

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

View Answer

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

View Answer

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

View Answer

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.

View Answer

The line $$y + 14 = 0$$ cuts the curve whose equation is $$x\left( {{x^2} + x + 1} \right) + y = 0$$ at

Releted MCQ Question on
Algebra >> Quadratic Equation

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

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

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

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

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Quadratic Equation

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The line $$y + 14 = 0$$ cuts the curve whose equation is $$x\left( {{x^2} + x + 1} \right) + y = 0$$ at

Releted MCQ Question on Algebra >> Quadratic Equation

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

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

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

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

Practice More Releted MCQ Question on Quadratic Equation

Practice More MCQ Question on Maths Section

Releted MCQ Question on
Algebra >> Quadratic Equation

Practice More Releted MCQ Question on
Quadratic Equation