You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quadratic Equation

Is the given ...

Question

Is the given equation quadratic ?
(2x + 1)(3x + 2) = 6(x - 1)(x - 2)

Open in App

Solution

$(2 x + 1) (3 x + 2) = 6 (x - 1) (x - 2)$
$\Rightarrow {6 x}^{2} + 7 x + 2 = 6 (x^{2} - 3 x + 2)$
$\Rightarrow {6 x}^{2} + 7 x + 2 = {6 x}^{2} - 18 x + 12$
$\Rightarrow 25 x = 10 \Rightarrow x = 0.4$ (Not quadratic equation)

Suggest Corrections

Similar questions

\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}, x \neq 0

Find the roots of the quadratic equation $6 x^{2} - 13 x + 6 = 0$ by using the quadratic formula.

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

x^{2} + \frac{1}{x^{2}} = 5

x - \frac{3}{x} = x^{2}

2 x^{2} - \sqrt{3 x} + 9 = 0

x^{2} - 2 x - \sqrt{x} - 5 = 0

x + \frac{1}{x} = 1

{(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4

x + \frac{1}{x} = x^{2}, x \neq 0

2 x^{2} + 6 x + \frac{9}{2} = 0

x^{2} - x + 3 = 0

2 x^{2} + \frac{5}{2} x - \sqrt{3} = 0

\sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0

\frac{1}{3} x^{2} + \frac{1}{5} x - 2 = 0

x^{2} - 3 x - \sqrt{x} + 4 = 0

x - \frac{6}{x} = 3

x + \frac{2}{x} = x^{2}

x^{2} - \frac{1}{x^{2}} = 5

{(x + 2)}^{3} = x^{3} - 8

(2 x + 3) (3 x + 2) = 6 (x - 1) (x - 2)

{(x + \frac{1}{x})}^{2} = 2 (x + \frac{1}{x}) + 3

Join BYJU'S Learning Program