Which of the Following Are Quadratic Equations?
Question:
Which of the following are quadratic equations?
(i) \(x^2+6x-4=0\)
(ii) \(\sqrt3x^2-2x+\frac12=0\)
(iii) \(x^2+\frac1{x^2}=5\)
(iv) \(x-\frac3x=x^2\)
(v) \(2x^2-\sqrt3x+9=0\)
(vi) \(x^2-2x-\sqrt x-5=0\)
(vii) \(3x^2-5x+9=x^2-7x+3\)
(viii) \(x+\frac1x=1\)
(ix) \(x^2-3x=0\)
(x) \(\left(x+\frac1x\right)^2=3\left(x+\frac1x\right)+4\)
(xi) \((2x+1)(3x+2)=6(x-1)(x-2)\)
(xii) \(x+\frac1x=x^2,\;x\ne0\)
(xiii) \(16x^2-3=(2x+5)(5x-3)\)
(xiv) \((x+2)^3=x^3-4\)
(xv) \(x(x+1)+8=(x+2)(x-2)\)
Solution
A quadratic equation can be written in the form
\[ ax^2+bx+c=0,\quad a\ne0 \]
(i) \(x^2+6x-4=0\) ⇒ Quadratic ✔
(ii) \(\sqrt3x^2-2x+\frac12=0\) ⇒ Quadratic ✔
(iii)
\[ x^2+\frac1{x^2}=5 \]
\[ x^4-5x^2+1=0 \]
Degree 4 ⇒ Not quadratic ✘
(iv)
\[ x-\frac3x=x^2 \]
\[ x^2-3=x^3 \]
\[ x^3-x^2+3=0 \]
Degree 3 ⇒ Not quadratic ✘
(v) \(2x^2-\sqrt3x+9=0\) ⇒ Quadratic ✔
(vi) \(x^2-2x-\sqrt x-5=0\)
Contains \(\sqrt x\) ⇒ Not quadratic ✘
(vii)
\[ 3x^2-5x+9=x^2-7x+3 \]
\[ 2x^2+2x+6=0 \]
Quadratic ✔
(viii)
\[ x+\frac1x=1 \]
\[ x^2+1=x \]
\[ x^2-x+1=0 \]
Quadratic ✔
(ix)
\[ x^2-3x=0 \]
Quadratic ✔
(x)
\[ \left(x+\frac1x\right)^2=3\left(x+\frac1x\right)+4 \]
\[ x^2+\frac1{x^2}+2=3x+\frac3x+4 \]
\[ x^4-3x^3-2x^2-3x+1=0 \]
Degree 4 ⇒ Not quadratic ✘
(xi)
\[ (2x+1)(3x+2)=6(x-1)(x-2) \]
\[ 6x^2+7x+2=6x^2-18x+12 \]
\[ 25x-10=0 \]
Linear equation ⇒ Not quadratic ✘
(xii)
\[ x+\frac1x=x^2 \]
\[ x^2+1=x^3 \]
\[ x^3-x^2-1=0 \]
Degree 3 ⇒ Not quadratic ✘
(xiii)
\[ 16x^2-3=(2x+5)(5x-3) \]
\[ 16x^2-3=10x^2+19x-15 \]
\[ 6x^2-19x+12=0 \]
Quadratic ✔
(xiv)
\[ (x+2)^3=x^3-4 \]
\[ x^3+6x^2+12x+8=x^3-4 \]
\[ 6x^2+12x+12=0 \]
Quadratic ✔
(xv)
\[ x(x+1)+8=(x+2)(x-2) \]
\[ x^2+x+8=x^2-4 \]
\[ x+12=0 \]
Linear equation ⇒ Not quadratic ✘
Answer
The quadratic equations are:
(i), (ii), (v), (vii), (viii), (ix), (xiii), (xiv)
Hence, the required quadratic equations are \[ \boxed{(i),(ii),(v),(vii),(viii),(ix),(xiii),(xiv)} \]