Question

Which of the following are quadratic equations in x ?

$v) \: \: x {}^{2} - 3x - \sqrt{x} + 4 = 0$

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


Class 10

Quadratic Equation

Answer 1

Hey there !!


Q. Which of the following are quadratic polynomial in x ?

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

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


▶ Solution :-

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

→ ${x}^{2} - 3x - \sqrt{x} + 4 = 0$ contains a term with $\sqrt{x}$ , i.e., ${x}^{ \frac{1}{2} }$ where $\frac{1}{2}$ is not a integer.

→ Therefore, it is not a quadratic polynomial.


•°• ${x}^{2} - 3x - \sqrt{x} + 4 = 0$ is not a quadratic equation.



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


$\bf \: = > {( \frac{ {x}^{2} + 1}{x} )}^{2} = 2( \frac{ {x}^{2} + 1}{x} ) + 3.$

$\bf \: = > \frac{ {( {x}^{2} + 1)}^{2} }{ {x}^{2} } = \frac{2 {x}^{2} + 2 }{x} + 3.$


$\bf \: = > { ( {x}^{2} + 1)}^{2} = {x} \cancel{^{2} } ( \frac{2 {x}^{2} + 2}{ \cancel{x}} ) + 3 \times {x}^{2} .$

=> ( x² + 1 )² = 2x( x² + 1 ) + 3x² .

=> x⁴ + 2x + 1 = 2x³ + 2x + 3x² .

=> x⁴ - 2x³ - 3x² + 2x - 2x + 1 = 0.

=> x⁴ - 2x³ - 3x² + 1 = 0.


▶This is not of the form ax² + bx + c = 0.

Hence, the given equation is not a quadratic equation.


✔✔ Hence, it is solved ✅✅.

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THANKS


#BeBrainly.

Answer 2

HEY THERE!!!


Question;

Which of the following are quadratic equations in x ?

=> x²-3x-√x+4=0


Method of Solution;

x²-3x-√x+4=0

Here, It Equation contains(Constant term) involving √x ,i.e. , x½ , Where ½ is not an integer.

Since, It not begin in the form of Quardratic Equation as ax²+bx+c=0.

Conclusion; x²-3x-√x+4=0 is not Quardratic Equation;

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


$\implies\tt\: (x + \frac{1}{x} ) {}^{2} = 2(x + \frac{1}{x} ) + 3 \\ \\ \\ \implies\sf{x}^{2} + \frac{1}{ {x}^{2}} + 2.x. \frac{1}{x} = 2x + \frac{2}{x} + 3 \\ \\ \\ \implies\sf{x}^{2} + \frac{1}{ {x}^{2}} + 2. \cancel{x}. \frac{1}{ \cancel{x}} = 2x + \frac{2}{x} + 3 \\ \\ \\ \\ \\ \implies\sf {x}^{2} + \frac{1}{ {x}^{2} } + 2 = \frac{ {2x}^{2} + 2 + 3x}{x} \\ \\ \\ \implies \frac{ {x}^{4} + 1 + {2x}^{2} }{ {x}^{2} } = \frac{ {2x}^{2} + 2 + 3x}{x} \\ \\ \: \: \implies\sf \: x( {x}^{4} + 1 + {2x}^{2} ) = {x}^{2} ( {2x}^{2} + 2 + 3x) \\ \\ \implies \sf{x}^{5} + x + {2x}^{3} = {2x}^{4} + {2x}^{2} + {3x}^{3} \\ \\ \implies\tt{x}^{5} - {2x}^{4} + {2x}^{3} - {3x}^{3} - {2x}^{2} + x = 0 \implies \\ \\ \sf \implies{x}^{5} - {2x}^{4} - {x}^{3} + {2x}^{2} + x = 0$


Here, This Equation is not in the form of ax²+bx+c=0 , Since It's Considered to be , It is not Quardratic Equation.



Thanks!!!