Question

which of the following is not a quadratic equation 1. 5x2-7x=0 2. -3x+9x2+8=0 3.√x+7/√x=4√x(x is not equal to 0) 4.1/3x+3x=4(x is not equal to 0)​

Answer 1

GIVEN :

The following equations are

i) $5x^2-7x=0$

ii) $-3x+9x^2+8=0$

iii) $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$)

iv) $\frac{1}{3x}+3x=4$  ( $x\neq 0$)

TO FIND :

Which of the given equation is not a quadratic equation.

SOLUTION :

Given equations are

i) $5x^2-7x=0$

ii) $-3x+9x^2+8=0$

iii) $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$)

iv) $\frac{1}{3x}+3x=4$  ( $x\neq 0$)

Now verify the equations which is not quadratic equation:

Quadratic equation is of the form

$ax^2+bx+c=0$ and its degree should be 2.

i) $5x^2-7x=0$

The equation $5x^2-7x=0$ is of the form of $ax^2+bx+c=0$ and its degree should be 2.

∴ $5x^2-7x=0$ is a quadratic equation.

ii) $-3x+9x^2+8=0$

The equation $-3x+9x^2+8=0$ is of the form of $ax^2+bx+c=0$ and its degree should be 2.

∴ $-3x+9x^2+8=0$ is a quadratic equation

iii) $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$)

Now solving $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$)

$\sqrt{x}+7=4\sqrt{x}\times \sqrt{x}$

$\sqrt{x}+7=4\sqrt{x}^2$

$\sqrt{x}+7=4x$

$\sqrt{x}+7-4x=0$

The equation $\sqrt{x}+7-4x=0$ is not in the form of $ax^2+bx+c=0$ and its not having degree 2.

iv) $\frac{1}{3x}+3x=4$  ( $x\neq 0$)

Now solving $\frac{1}{3x}+3x=4$

$\frac{1+9x^2}{3x}=4$

$1+9x^2=12x$

$1+9x^2-12x=0$

$9x^2-12x+1=0$

The equation $9x^2-12x+1=0$ is of the form of $ax^2+bx+c=0$ and its degree should be 2.

∴ option iii) $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$) is not a quadratic equation.

∴ option iii) $\frac{\sqrt{x}+7}{\sqrt{x}}=4\sqrt{x}$ ( $x\neq 0$) is correct.