Question

(x - 2)(x + 1) = (x - 1)(x + 3) check whether it is quadratic equation or not ​

Answer 1

If the equation will have the degree 2,then the equation will be considered as quadratic equation.....

${x - 2} \times {x + 1} = x - 1 \times x + 3 \\ x \times {$

Answer 2

Answer:

$\Large\boxed{\mathsf{X=\frac{1}{3}=0.33 }}$

Step-by-step explanation:

GIVEN (QUESTION):

(x - 2)(x + 1) = (x - 1)(x + 3) check whether it is quadratic equation or not ​

TO FIND:

The quadratic equation of (x-2)(x+1)=(x-1)(x+3)

TO SOLVE:

With FOIL method.

SOLUTIONS:

First, expand by using with FOIL method.

$\displaystyle \mathsf{FOIL \quad METHOD}$

$\displaystyle \mathsf{(A+B)(C+D)=AC+AD+BC+BD}}$

$\displaystyle \mathsf{A=X}\\\\\displaystyle \mathsf{B=(-2)}\\\\\displaystyle \mathsf{C=X}\\\\\displaystyle \mathsf{D=1}$

$\displaystyle \mathsf{XX+X*1+(-2)X+1(-2)*1}$

$\displaystyle \mathsf{XX+1*\:X-2x-2* \:1}}}$

Solve.

Add similar elements.

$\displaystyle \mathsf{1*x-2x=-x}}$

$\displaystyle \mathsf{XX-X-2*1}}}$

$\displaystyle \mathsf{X^2=XX}$

Multiply.

$\displaystyle \mathsf{2\times1=2}$

Then, rewrite the equation problem.

$\displaystyle \mathsf{X^2-X-2}}$

Expand.

(x-1)(x+3)

A=X

B=(-1)

C=X

D=3

$\displaystyle \mathsf{(x-1)(x+3)=x^2+2x-3}$

$\displaystyle \mathsf{x^2-x-2=x^2+2x-3}}$

Add by 2 from both sides.

$\displaystyle \mathsf{x^2-x-2+2=x^2+2x-3+2}$

Solve.

$\displaystyle \mathsf{x^2-x=x^2+2x-1}$

Then, you subtract by x²+2x from both sides.

$\displaystyle \mathsf{x^2-x-\left(x^2+2x\right)=x^2+2x-1-\left(x^2+2x\right)}}}$

Solve.

$\displaystyle \mathsf{-3x=-1}}$

Divide by -3 from both sides.

$\displaystyle \mathsf{\frac{-3x}{-3}=\frac{-1}{-3}}$

Solve.

Change negative sign to positive sign.

$\displaystyle \mathsf{\boxed{\mathsf{X=\frac{1}{3}=0.33}}}}}} }}}$

$\Large\boxed{\longrightarrow \mathsf{X=\frac{1}{3}=0.33}}}}}$

It is not quadratic equation. So, the correct answer is x=1/3=0.33.