Check whether the following are quadratic equations:
$(x + 1) (x - 1) = (x + 2) (x + 3)$

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Solution

Given, $(x + 1) (x - 1) = (x + 2) (x + 3)$
$\Rightarrow x^{2} - 1^{2} = x^{2} + 5 x + 6$
$\Rightarrow x^{2} - 1^{2} - x^{2} - 5 x - 6 = 0$
$\Rightarrow - 5 x - 7 = 0$
$∵$ The highest power of x in the equation is 1;
$∴$ It is not a quadratic equation.

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Similar questions

Check whether the following are quadratic equations :
(i) ${(x + 1)}^{2} = 2 (x - 3)$
(ii) $x^{2} - 2 x = (- 2) (3 - x)$
(iii) $(x - 2) (x + 1) = (x - 1) (x + 3)$
(iv) $(x - 3) (2 x + 1) = x (x + 5)$
(v) $(2 x - 1) (x - 3) = (x + 5) (x - 1)$
(vi) $x^{2} + 3 x + 1 = {(x - 2)}^{2}$
(vii) ${(x + 2)}^{3} = 2 x (x^{2} - 1)$
(viii) $x^{3} - 4 x^{2} - x + 1 = {(x - 2)}^{3}$

Q. Question 1(iii)
Check whether the following are quadratic equations:
\((iii) (x -2)(x + 1) = (x - 1)(x + 3)

Q. Check whether the following are quadratic equations:

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

Q. Check whether the following is quadratic equation.

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

Q. Check whether the following are Quadratic equations

(2 x - 1) (x - 3) = (x + 5) (x - 1)

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Check whether the following are quadratic equations:(x+1)(x−1)=(x+2)(x+3)

Given,(x+1)(x−1)=(x+2)(x+3)⇒x2−12=x2+5x+6⇒x2−12−x2−5x−6=0⇒−5x−7=0 ∵ The highest power of x in the equation is 1;∴ It is not a quadratic equation.

Check whether the following are quadratic equations:
$(x + 1) (x - 1) = (x + 2) (x + 3)$

Given, $(x + 1) (x - 1) = (x + 2) (x + 3)$
$\Rightarrow x^{2} - 1^{2} = x^{2} + 5 x + 6$
$\Rightarrow x^{2} - 1^{2} - x^{2} - 5 x - 6 = 0$
$\Rightarrow - 5 x - 7 = 0$
$∵$ The highest power of x in the equation is 1;
$∴$ It is not a quadratic equation.