In each of these question two equations I & II with variables a & b are given You have to solve both the equations to find the values of a & b
Mark answer if
a) a
We can easily solve equation I to get a = -1
But we cannot solve $b^{2}=\pm4$. Square root of negative number is not a real number.
Hence, we cannot find a value of b. Therefore, we cannot establish a relationship between a and b.
We can easily solve equation I to get a = -1
But we cannot solve $b^{2}=\pm4$. Square root of negative number is not a real number.
Hence, we cannot find a value of b. Therefore, we cannot establish a relationship between a and b.
2. If x+y+z = 0 , then what is the value of $\frac{x^{2}}{3z}+\frac{y^{2}}{3xz}+\frac{z^{2}}{3x}$
4. The sum of a number and 4 times its reciprocal is 4. Find the number.
5. Which of the following equations has the sum of its roots as 5?
7. If 4(2x + 3) > 5 x and 5x 3(2x7) > 3x 1, then x can take which of the following values?
10. Simplify $\frac{(b^{5}x^{2}a ^{3}z^{4})*(b^{3}x^{2}a^{4}z^{5} )}{(a^{2}b^{3}z^{2})}$