Question

1. Find two different solutions of 3x+y=11

2. Find the remainder when x³+2x²+4x+3 is divided by the polynomial x-1

3. Insert three rational numbers between 1/4 and 7/8

4. Check whether (x-2) is a factor of x³-2x²+3x-6

5. If √2=1.4142....check whether 3+√2 is rational or irrational ? Give reason.

6. Write two different linear equations in two variables.

7. Evaluate the value of (101)² by using suitable identity

8. Find the value of k, if x=3, y=2 is a solution of the equation 3x+4y=k. Find two more solutions of the resultant equation

9. Find the product of (6x-y+z) (6x-y+z)

10. If 3 is the zero of the polynomial p(x)= 2x²+3x+7a then find the value of 'a'.

Answer 1

X=3,Y=2 are you read in class 9

Answer 2

Answer:

1) 3x+y = 11

Put x =1, y=8

Put x=2, y=5

Any two solutions: (1,8) and (2,5)

2) Remainder = 7x+3

3) 1/4 and 7/8

2/8 and 7/8

Three rational numbers = $\frac{3}{8}, \frac{4}{8}, \frac{5}{8}$

4) f(x) = x³-2x²+3x-6

If f(2) = 0, then x-2 is a factor

f(2) = $2^{3}-2×2^{2}+3(2)-6$ = 0

Therefore, x-2 is a factor.

5) 3 +√2 = 4.4142...

It is irrational since value is non terminating and recurring.

6) Two different linear equations in two variables say x and y.

x+y = 3

2x-6y = 8

7) (101)^{2} =$(100+1)^{2}$  [Using identity (a+b)^{2}]

                = $(100)^{2}$ +2(100)(1)+1

                = 10201

8) Putting x = 3 and y=2 in  3x+4y=k

3(3)+4(2)= k

k =14

9) (6x-y+z)(6x-y+z) = (6x-y+z)^{2}  

Using identity $(a+b+c)^{2} = a^{2}+ b^{2}+ c^{2}+2xy+2yz+2zx$

(6x-y+z)^{2}  = $6x^{2}+ y^{2}+ z^{2}+2(6x)y+2(-y)z+2z(6x)$

                    = $36x^{2}+ y^{2}+ z^{2}-12xy+-2yz+12xz)$

10) p(x) = $2x^{2} +3x+7a$

3 is zero, p(3)= 0

$23^{2} +3(3)+7a$ = 0

27+7a = 0 ⇒ a = $\frac{-27}{7}$