Q.1 Which of the following is not a quadratic equation?

1. $2(\mathrm{x}-1{)}^2=4{\mathrm{x}}^2-2\mathrm{x}+1$
2. $2\mathrm{x}-{\mathrm{x}}^2={\mathrm{x}}^2+5$
3. $(\sqrt{2}\mathrm{x}+\sqrt{3}{)}^2+{\mathrm{x}}^2=3{\mathrm{x}}^2-5\mathrm{x}$
4. ${\left({\mathrm{x}}^2+2\mathrm{x}\right)}^2={\mathrm{x}}^4+3+4{\mathrm{x}}^3$

(OR)

Q.2 The sum of first five positive integers divisible by 6 is :

1. 180
2. 90
3. 45
4. 30

Answer : 3

Four equations are given and we need to check which of them is not a quadratic equation.

It is known that a quadratic equation is an equation that has the highest power of 2, or in other words, have the highest power term as ax², where a is a constant.

Equation 1: $2(\mathrm{x}-1{)}^2=4{\mathrm{x}}^2-2\mathrm{x}+1$: On simplifying, we get the highest power term of the equation as x². Therefore, it is a quadratic equation.

Equation 2: $2\mathrm{x}-{\mathrm{x}}^2={\mathrm{x}}^2+5$: On simplifying, we get the highest power term of the equation as ax². Therefore, it is a quadratic equation.

Equation 3: $(\sqrt{2}\mathrm{x}+\sqrt{3}{)}^2+{\mathrm{x}}^2=3{\mathrm{x}}^2-5\mathrm{x}$: On simplifying, we get the highest power term of the equation as x. Therefore, it is not a quadratic equation.

Equation 4: ${\left({\mathrm{x}}^2+2\mathrm{x}\right)}^2={\mathrm{x}}^4+3+4{\mathrm{x}}^3$: On simplifying, we get the highest power term of the equation as x². Therefore, it is a quadratic equation.

Hence, $(\sqrt{2}\mathrm{x}+\sqrt{3}{)}^2+{\mathrm{x}}^2=3{\mathrm{x}}^2-5\mathrm{x}$ is not a quadratic equation.

(OR)

Answer : 2

We need to find the sum of first five positive integers divisible by 6.

From the given data, the AP having a common difference as 6 can be framed as 6, 12, 18, .....

It is known that the sum of an AP is given by

${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right]$, where S_n is the sum of terms, n is the number of terms, a is the first term and d is the common difference of AP.

On substituting the values, we get the sum as

$\begin{array}{l}\Rightarrow {\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2\mathrm{a}+(\mathrm{n}-1\left)\mathrm{d}\right]\\ \Rightarrow {\mathrm{S}}_5=\frac{5}{2}\left[2\right(0)+(5-1\left)0\right]\\ \Rightarrow {\mathrm{S}}_5=\frac{5}{2}[12+(4\left)0\right]\\ \Rightarrow {\mathrm{S}}_5=\frac{5}{2}[12+24]\\ \Rightarrow {\mathrm{S}}_5=\frac{5}{2}\times 36\\ \Rightarrow {\mathrm{S}}_5=90\end{array}$

Hence, the sum of first five positive integers divisible by 6 is 90.