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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\left(x^{2}-1\right) $
(viii) $ x^{3}-4 x^{2}-x+1=(x-2)^{3} $
To do:
We have to check whether the given equations are quadratic.
Solution:
The standard form of a quadratic equation is $ax^2+bx+c=0$.
Therefore,
(i) $(x+ 1)^2=2(x-3)$
$x^2+2(x)(1)+(1)^2 =2x-6$
$x^2+2x-2x+1+6=0$
$x^2+7=0$
$x^2+0x+7=0$ is of the form $ax^2+bx+c=0$
Therefore, $(x+ 1)^2=2(x-3)$ is a quadratic equation.
(ii) $x^2 - 2x = (- 2) (3-x)$
$x^2-2x=-6+2x$
$x^2-2x-2x+6=0$
$x^2+-4x+6=0$
$x^2-4x+6=0$ is of the form $ax^2+bx+c=0$
Therefore, $x^2 - 2x = (- 2) (3-x)$ is a quadratic equation.
(iii) $(x – 2) (x + 1) = (x – 1) (x + 3)$
$x(x+1)-2(x+1)=x(x+3)-1(x+3)$
$x^2+x-2x-2=x^2+3x-x-3$
$x^2-x^2-x-2x-2+3=0$
$-3x+1=0$ is not of the form $ax^2+bx+c=0$
Therefore, $(x – 2) (x + 1) = (x – 1) (x + 3)$ is not a quadratic equation.
(iv) $(x – 3) (2x + 1) = x (x + 5)$
$x(2x+1)-3(2x+1)=x(x)+x(5)$
$2x^2+x-6x-3=x^2+5x$
$2x^2-x^2-5x-5x-3=0$
$x^2-10x-3=0$ is of the form $ax^2+bx+c=0$
Therefore, $(x – 3) (2x + 1) = x (x + 5)$ is a quadratic equation.
(v) $(2x – 1) (x – 3) = (x + 5) (x – 1)$
$2x(x-3)-1(x-3)=x(x-1)+5(x-1)$
$2x^2-6x-x+3=x^2-x+5x-5$
$2x^2-x^2-7x-4x+3+5=0$
$x^2-11x+8=0$ is of the form $ax^2+bx+c=0$
Therefore, $(2x – 1) (x – 3) = (x + 5) (x – 1)$ is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x – 2)^2$
$x^2 + 3x + 1 = x^2 – 2(x)(2)+(2)^2$
$x^2 + 3x + 1 = x^2-4x+4$
$x^2-x^2+3x+4x+1-4=0$
$7x-3=0$ is not of the form $ax^2+bx+c=0$
Therefore, $x^2 + 3x + 1 = (x – 2)^2$ is not a quadratic equation.
(vii) $(x + 2)^3 = 2x(x^2 – 1)$
$x^3 + 2^3 + 3(x)(2) (x + 2) = 2x^3 - 2x$
$x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x$
$2x^3-x^3 - 6x^2 - 2x -12x - 8 = 0$
$x^3-6x^2-14x-8=0$ is not of the form $ax^2+bx+c=0$
Therefore, $(x + 2)^3 = 2x(x^2 – 1)$ is not a quadratic equation.
(viii) $x^3 -4x^2 -x + 1 = (x-2)^3$
$x^3 - 4x^2 - x + 1 = x^3-2^3 + 3(x)(-2)(x - 2)$
$x^3 - 4x^2 -x + 1 = x^3 - 6x^2 + 12x - 8$
$x^3-x^3-4x^2+6x^2 - x-12x + 1+8 = 0$
$2x^2-13x+9=0$ is of the form $ax^2+bx+c=0$
Therefore, $x^3 -4x^2 -x + 1 = (x-2)^3$ is a quadratic equation.
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