CBSE Class 10 Maths Important Questions

Solution

Assume $5+2\sqrt{3}=r$ for a rational number $r$. Then $\sqrt{3}=(r-5)/2$, which is rational. This contradicts the irrationality of $\sqrt{3}$. Therefore $5+2\sqrt{3}$ is irrational.

Solution

Assume $p+q\sqrt{5}=r$ for a rational number $r$. Since $q\ne 0$, $\sqrt{5}=(r-p)/q$, which is rational. This contradicts the irrationality of $\sqrt{5}$. Hence $p+q\sqrt{5}$ is irrational.

Solution

The exponents of $2$ and $5$ are odd. Multiplying by $2\times 5=10$ gives $2^4\times 3^2\times 5^2$, whose every prime exponent is even.

Solution

Exponents must become multiples of $3$. The least corrections are $2^2$, $3^1$ and $5^2$. Thus the multiplier is $2^2\times 3\times 5^2=150$, producing $2^6\times 3^3\times 5^3$.

Solution

Suppose $\sqrt{2}=a/b$ in lowest terms, where $a,b$ are positive integers. Then $a^2=2b^2$, so $a$ is even; write $a=2k$. Hence $b^2=2k^2$, so $b$ is even. This contradicts that $a/b$ is in lowest terms. Therefore $\sqrt{2}$ is irrational.

Solution

Assume $\sqrt{3}=a/b$ in lowest terms. Then $a^2=3b^2$, so $3$ divides $a$; write $a=3k$. Substitution gives $b^2=3k^2$, so $3$ divides $b$. This contradicts lowest terms. Thus $\sqrt{3}$ is irrational.

Solution

Suppose $7-3\sqrt{5}=r$, where $r$ is rational. Then $\sqrt{5}=(7-r)/3$, which would be rational, contradicting the irrationality of $\sqrt{5}$. Hence the expression is irrational.

Solution

Since $\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$, the expression equals $3\sqrt{2}$. If it were rational, then dividing by the nonzero rational number $3$ would make $\sqrt{2}$ rational, a contradiction. Therefore the expression is irrational.

Solution

For $a{x}^2$ + bx + c, product of zeroes is c/a. Here a = 2 and c = 3, so alpha beta = 3/2.

Solution

D = $b^2$ - 4ac = (-$6{)}^2$ - 4(1)(9) = 36 - 36 = 0. When D = 0, roots are real and equal.