Which of the following are quadratic equations- i x2-6x-4 - 0- ii -3x2-2x- 1-2 - 0- iii x2-1-x2 - 5- iv x- 3-x - x2- v 2x2- -3x-9 - 0- vi x2-2x - -x - 5 - 0- vii 3x2-5x-9 - x2-7x-3- viii x - 1-x - 1- ix x2-3x - 0- x x- 1-x 2 - 3 x-1-x -4- xi 2x-13x-2 - 6x-1x-2- xii x - 1-x - x2- x - 0- xiii 16x2-3 - 2x-5 5x-3 - xivx-23 - x3 - 4- xv x x-1-8 - x - 2 x-2

A quadratic equation is of the form ax^2 + bx +c = 0 . and max power of x should be 2. ie (i)Yes (ii)NO, because there is no term of x^2 (iii)NO, after cros ...

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Byju's Answer

Standard X

Mathematics

Quadratic Equations

Which of the ...

Question

Which of the following are quadratic equations?
$(i) x^{2} + 6 x - 4 = 0$
( $i i) \sqrt{3 x^{2}} - 2 x + \frac{1}{2} = 0$
$(i i i) x^{2} + \frac{1}{x^{2}} = 5$
$(i v) x - \frac{3}{x} = x^{2}$
$(v) 2 x^{2} - \sqrt{3 x} + 9 = 0$
$(v i) x^{2} - 2 x - \sqrt{x} - 5 = 0$
$(v i i) 3 x^{2} - 5 x + 9 = x^{2} - 7 x + 3$
$(v i i i) x + \frac{1}{x} = 1$
$(i x) x^{2} - 3 x = 0$
$(x) {(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$
$(x i) (2 x + 1) (3 x + 2) = 6 (x - 1) (x - 2)$
$(x i i) x + \frac{1}{x} = x^{2}, x \neq 0$
$(x i i i) 16 x^{2} - 3 = (2 x + 5) (5 x - 3)$
$(x i v) (x + 2)^{3} = x^{3} - 4$
$(x v) x (x + 1) + 8 = (x \div 2) (x - 2)$

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Solution

A quadratic equation is of the form $a x^{2} + b x + c = 0$ and maximum power of $x$ should be 2. i.e $a \neq 0$
The power of the variable should not be negative, in fraction, in square root.

(i) Yes
(ii) $\sqrt{3 x^{2}} - 2 x + \frac{1}{2} = 0$
The above equation will become $\sqrt{3} x - 2 x + \frac{1}{2} = 0$
No,the given equation is not a quadratic equation because there is no term of $x^{2}$
(iii) $x^{2} + \frac{1}{x^{2}} = 5$
The above equation will become $x^{4} - 5 x^{2} + 1 = 0$
No, after cross multiplying we'll see that the maximum power of $x$ is $4$ .
(iv) $x - \frac{3}{x} = x^{2}$
The above equation will become after simplification $x^{3} - x^{2} + 3 = 0$
No, maximum power is $3$ that's why the given equation is not a quadratic one.
(v) $2 x^{2} - \sqrt{3 x} + 9 = 0$
The above equation will become after simplification $2 x^{2} - 3^{\frac{1}{2}} x^{\frac{1}{2}} + 9 = 0$
No,as there is a term of $x^{\frac{1}{2}}$
(vi) No, because there is a term with $\sqrt{x}$ .
(vii) $3 x^{2} - 5 x + 9 = x^{2} - 7 x + 3$
The above equation will become after simplification $2 x^{2} + 2 x + 6 = 0$
Yes, the given equation is a quadratic one.
(viii) $x + \frac{1}{x} = 1$
The above equation will become after simplification $x^{2} - x + 1 = 0$
Yes, the given equation is a quadratic one.
(ix) Yes
(x) ${(x + \frac{1}{x})}^{2} = 3 (x + \frac{1}{x}) + 4$
The above equation will become after simplification $x^{4} - 3 x^{3} - 2 x^{2} - 3 x + 1 = 0$
No, maximum power of $x$ will be $4$ .
(xi) $(2 x + 1) (3 x + 2) = 6 (x - 1) (x - 2)$
The above equation will become after simplification $25 x - 10 = 0$
No, because when we simplify the equation there will be no term of $x^{2}$
(xii) $x + \frac{1}{x} = x^{2}$
The above equation will become after simplification $x^{3} - x^{2} - 1 = 0$
No, maximum power is $3$ , so the given equation is not a quadratic one.
(xiii) $16 x^{2} - 3 = (2 x + 5) (5 x - 3)$
The above equation will become after simplification $2 x^{2} - 3 x + 4 = 0$
Yes, the given equation is a quadratic one.
(xiv) $(x + 2)^{3} = x^{3} - 4$
The above equation will become after simplification $x^{2} + 2 x + 2 = 0$
Yes, when we simplify the equation will not have $x^{3}$ term.
(xv) $x (x + 1) + 8 = (x \div 2) (x - 2)$
The above equation will become after simplification $x^{2} + 4 x + 16 = 0$
Yes, the given equation is a quadratic one.

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Similar questions

Q. Which of the following are quadratic equations?

(i) x² + 6x − 4 = 0
(ii)

\sqrt{3} x^{2} - 2 x + \frac{1}{2} = 0

(iii)

x^{2} + \frac{1}{x^{2}} = 5

(iv)

x - \frac{3}{x} = x^{2}

(v)

2 x^{2} - \sqrt{3 x} + 9 = 0

(vi)

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

(vii) 3x² − 5x + 9 = x² − 7x + 3
(viii)

x + \frac{1}{x} = 1

(ix) x² − 3x = 0
(x)

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

(xi) (2x + 1) (3x + 2) = 6(x − 1) (x − 2)
(xii)

x + \frac{1}{x} = x^{2}, x \neq 0

(xiii) 16x² − 3 = (2x + 5) (5x − 3)
(xiv) (x + 2)³ = x³ − 4
(xv) x(x + 1) + 8 = (x + 2) (x − 2)

In the frollowing determine whetehr the given quadratic equations have real roots
$(i) 16 x^{2} = 24 x + 1 (i i) x^{2} + x + 2 = 0 (i i i) \sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0 (i v) 3 x^{2} - 2 x + 2 = 0 (v) 2 x^{2} - 2 \sqrt{3} x + 3 = 0 (v i) 3 a^{2} x^{2} + 8 a b x + 4 b^{2} = 0, a \neq 0 (v i i) 3 x^{2} + 1 \sqrt{3} x - 5 = 0 (v i i i) x^{2} - 2 x + 1 = 0 (i x) 2 x^{2} + 5 \sqrt{3} x + 6 = 0 (x) \sqrt{2} x^{2} + 7 x + 5 \sqrt{2} = 0 (x i) 2 x^{2} - 2 \sqrt{2} x + 1 = 0 (x i i) 3 x^{2} - 5 x + 2 = 0$

Solve :

$(i) \frac{1}{3} x - 6 = \frac{5}{2} (i i) \frac{2 x}{3} - \frac{3 x}{8} = \frac{7}{12} (i i i) (x + 2) (x + 3) + (x - 3) (x - 2) - 2 x (x + 1) = 0 (i v) \frac{1}{10} - \frac{7}{x} = 35 (v) 13 (x - 4) - 3 (x - 9) - 4 (x + 4) = 0 (v i) x + 7 - \frac{8 x}{3} = \frac{17 x}{6} - \frac{5 x}{8} (v i i) \frac{3 x - 2}{4} - \frac{2 x + 3}{3} = \frac{2}{3} - x (v i i i) \frac{x + 2}{6} - (\frac{11 - x}{3} - \frac{1}{4}) = \frac{3 x - 4}{12} (i x) \frac{2}{5 x} - \frac{5}{3 x} = \frac{1}{15} (x) \frac{x + 2}{3} - \frac{x + 1}{5} = \frac{x - 3}{4} - 1 (x i) \frac{3 x - 2}{3} + \frac{2 x + 3}{2} = x + \frac{7}{6} (x i i) x - \frac{x - 1}{2} = 1 - \frac{x - 2}{3} (x i i i) \frac{9 x + 7}{2} - (x - \frac{x - 2}{7}) = 36 (x i v) \frac{6 x + 1}{2} + 1 = \frac{7 x - 3}{3}$