Question

Which of the following are quadratic equations?
(x)$(x+\frac{1}{x})^{2}=3(x+\frac{1}{x})+4$
(xi)(2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)x+1x=x2, x≠0

Answer 1

SOLUTION  :

(x) Given : (x + 1/x)² = 3(x + 1/x) + 4

[(x² + 1)/x]² = 3[(x² + 1)/x] + 4

[((x²)² + 1² + 2x²)/x²] = (3x² + 3)/x + 4

[(a + b)² = a² + b² + 2ab]

(x⁴ + 1 + 2x²)/x² = (3x² + 3+ 4x)/x

(x⁴ + 1 + 2x²) = x² (3x² + 3+ 4x)/x

(x⁴ + 1 + 2x²) = x (3x² + 3+ 4x)

(x⁴ + 1 + 2x²) = 3x³ + 3x + 4x²

x⁴  - 3x³ + 2x² - 4x² - 3x + 1  = 0

x⁴ - 3x³ - 2x² - 3x +1 = 0

It is not of the form ax² + bx + c = 0, a ≠ 0,because the degree of the equation is of 4.

Hence, the above given equation does not represent a quadratic equation.

(xi) Given : (2x + 1)(3x + 2) = 6(x -1)(x - 2)

6x² + 4x + 3x + 2  = (6x - 6)(x - 2)

6x² + 4x + 3x + 2  = 6x² - 12x - 6x + 12

6x² + 7x + 2  = 6x² - 18x  + 12

6x² - 6x² + 7x +18x  + 2 - 12 = 0

25x - 10 = 0

5(5x - 2) = 0

5x - 2 = 0

The above equation is not of the form ax² + bx + c = 0, a ≠ 0,because the degree of the equation is of 1 (linear equation).

Hence, the above given equation does not represent a quadratic equation.

(xii) Given : x + 1/x = x²

(x² + 1)/x = x²

x² + 1 = x² × x

x² + 1 = x³

-x³ + x² + 1 = 0

The above equation is not of the form ax² + bx + c = 0, a ≠ 0,because the degree of the equation is of 3.

Hence, the above given equation does not represent a quadratic equation.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Which of the following are quadratic equations?

(x)$(x+\frac{1}{x})^{2}=3(x+\frac{1}{x})+4$√√ is answer

(xi)(2x + 1) (3x + 2) = 6(x − 1) (x − 2)

(xii)x+1x=x2, x≠0