The degree of the equation, given by $(x + 2) (x - 1) = (x + 1) (x + 3)$ , is

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $1$
LHS $\Rightarrow (x + 2) (x + 1)$
$\Rightarrow x^{2} + x - 2$ ....(1)
RHS $\Rightarrow (x + 1) (x + 3)$
$\Rightarrow x^{2} + 2 x + 3$ ....(2)

$∴ x^{2} + x - 2 = x^{2} + 2 x + 3$
$\Rightarrow x - 2 = 2 x + 3$
$\Rightarrow x + 5 = 0$

Therefore, its an equation of degree 1.

Hence, option C.

Suggest Corrections

Similar questions

Solve the following quadratic equation by factorization.
$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

Q. Solve the following quadratic equation by factorization, the sum of the roots are :

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

The equation $(x - 2) (x + 1) = (x - 1) (x + 3)$ is a quadratic equation.

Q. Solve the following equation by using the general formula , if the equation has a solution R :
i)

\frac{1}{x} - \frac{1}{x - 3} = 3, x \neq 0, 2

ii)

x + \frac{1}{x} = 3, x \neq 0

iii)

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

iv)

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

\frac{1}{x + 1} + \frac{2}{x + 2} = \frac{4}{x + 4}; x \neq 1, - 2, - 4

vi)

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

Q. Find the degree of the equation : x² – 2x + 1 = x² – 3.

Join BYJU'S Learning Program

The degree of the equation, given by (x+2)(x−1)=(x+1)(x+3), is

The correct option is D 1LHS ⇒(x+2)(x+1)⇒x2+x−2 ....(1)RHS ⇒(x+1)(x+3)⇒x2+2x+3 ....(2)∴x2+x−2=x2+2x+3⇒x−2=2x+3⇒x+5=0Therefore, its an equation of degree 1.Hence, option C.

The degree of the equation, given by $(x + 2) (x - 1) = (x + 1) (x + 3)$ , is