Question

Check whether (x+1)^2 =2(x+3) is quadratic equation or not?​

Answer 1

Answer:

(x+1)^2 = x^2 + 2x + 1

x^2 + 2x + 1 = 2x + 6

x^2 + 2x -2x + 1 - 6 = 0

x^2 -5 = 0

x^2 = 5

Therefore, it is not a quadratic equation.

Answer 2

$\large\sf\underline{Given\::}$

Equation :

• $\sf\:(x+1) ^{2}=2(x+3)$

‎

$\large\sf\underline{To\::}$

• Check whether the equation is quadratic Or not .

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$\large\sf\underline{Criteria\:to\:be\:quadratic\::}$

• Quadratic equation are generally in the form $\sf\:ax^{2}+bx+c=0$

where a ≠ 0 and a, b c are real numbers

• Highest power in the quadratic equation is 2

So now let's check if the equation is quadratic.

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$\large\sf\underline{Solution\::}$

$\sf\:(x+1) ^{2}=2(x+3)$

Using an identity :

• $\large{\mathfrak{(a+b)^{2}=a^{2}+2ab+b^{2}}}$

Here in our question :

• a = x and b = 1

$\sf\implies\:x^{2}+2 \times x \times 1+1^{2}=2(x+3)$

$\sf\implies\:x^{2}+2x+1=2(x+3)$

• Multiplying the terms in RHS and removing the brackets

$\sf\implies\:x^{2}+2x+1=2x+6$

• Now transposing +2x from RHS to LHS it becomes -2x

$\sf\implies\:x^{2}+2x-2x+1=6$

• Like terms with opposite signs gets cancelled

$\sf\implies\:x^{2}+\cancel{2x}-\cancel{2x}+1=6$

$\sf\implies\:x^{2}+1=6$

• Now transposing +6 from RHS to LHS it becomes -6

$\sf\implies\:x^{2}+1-6=0$

$\small{\underline{\boxed{\mathrm\red{\implies\:x^{2}-5\:=\:0}}}}$ ★

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So we simplified the given equation in $\sf\:x^{2}-5$ . Let's verify if this simplified equation follows the criteria of quadratic equations.

So in the simplified equation :

• a = 1

• b = 0

• c = -5

Therefore the first criteria is followed which states that a can't be zero whereas b and c might be zero.

The highest power in the equation is 2 . That means it follows the second criteria too. So the given equation is quadratic.

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$\dag\:\underline{\sf \:The\:given\:equation\:is\:quadratic\:.}$‎

‎!! Hope it helps !!