Algebra Additional Questions-10th Std Maths-Book Back Question And Answer
I. Choose the correct answer.
Question 1.
The HCF of x2 – y2; x3 – y3, …………. xn – yn where n ∈ N is
(1) x – y
(2) x + y
(3) xn – yn
(4) do not intersect
Answer:
(1) x – y
Question 2.
Which of the following is correct.
(i) Every polynomial has finite number of multiples
(ii) LCM of two polynomials of degree “2” may be a constant
(iii) HCF of 2 polynomials may be a constant
(iv) Degree of HCF of two polynomials is always less than degree of L.C.M.
(1) (i) and (iii)
(2) (iii) and (iv)
(3) (iii) only
(4) (iv) only
Answer:
(3) (iii) only
Question 3.
The HCF of x2 – 2xy + y2 and x4 – y4 is …………….
(1) 1
(2) x + y
(3) x – y
(4) x2 – y2
Answer:
(3) x – y
Question 4.
The L.C.M. of ak ak+3, ak+5 where K ∈ N is …………
(1) ak+5
(2) ak
(3) ak+6
(4) ak+9
Answer:
(1) ak+5
Question 5.
The LCM of (x + 1)2 (x – 3) and
(x2 – 9) (x + 1) is
(1) (x + 1)3 (x2 – 9)
(2) (x + 1)2 x2 – 9)
(3) (x + 1)2 (x – 3)
(4) (x – 9) (x + 1)
Answer:
(2) (x + 1)2(x2 – 9)
Question 6.
If a3a−b is added with b3b−a then the new expressions is …………
(1) a2 – ab + b2
(2) a2 + ab + b2
(3) a3 + b3
(4) a3 – b3
Answer:
(2) a2 + ab + b2
Question 7.
The solution set of x + 1x = 52 is ………….
(1) 2,12
(2) 2,-12
(3) -2, – 12
(4) -2, 72
Answer:
(1) 2,12
Question 8.
On dividing x2−25x+3 by x+5x2−9 is equal to ……………….
(1) (x – 5)(x + 3)
(2) (x + 5) (x – 3)
(3) (x – 5)(x – 3)
(4) (x + 5)(x + 3)
Answer:
(3) (x – 5)(x – 3)
Question 9.
The square root of (x + 11)2 – 44x is ………….
(1)|(x – 11)2
(2) |x + 11|
(3) |11 – x2|
(4) |x – 11|
Answer:
(4) |x – 11|
Question 10.
If α, β are the zeros of the polynomial p(x) = 4x2 + 3x + 7 then 1α + 1β is equal to …………
(1) 73
(2) – 73
(3) 37
(4) – 37
Answer:
(4) – 37
Question 11.
The value of 
……….
(1) -5
(2) 5
(3) 4
(4) -3
Answer:
(2) 5
Question 12.
If α and β are the roots of the equation ax2 + bx + c = 0 then (α + β)2 is ……………..
(1) −b2a2
(2) −c2a2
(3) −b2a2
(4) bca
Answer:
(3) −b2a2
Question 13.
The roots of the equation x2 – 8x + 12 = 0 are
(1) real and equal
(2) real and rational
(3) real and irrational
(4) unreal
Answer:
(2) real and rational
Question 14.
If one root of the equation is the reciprocal of the other root in ax2 + bx + c = 0 then …………
(1) a = c
(2) a = b
(3) b = c
(4) c = 0
Answer:
(1) a = c
Question 15.
If α and β are the roots of the equation x2 + 2x + 8 = 0 then the value of αβ+βα is ………………
(1) 12
(2) 6
(3) 32
(4) –32
Answer:
(4) –32
Question 16.
are………
(1) 4, 6, 6
(2) 6, 6, 4
(3) 6, 4, 6
(4) 4, 4, 6
Answer:
(3) 6, 4, 6
Question 17.
If [-1 -2 4] 
then the value of “a” is ………….
(1) 2
(2) -4
(3) 4
(4) -2
Answer:
(4) -2
Question 18.
The matrix A given by (aij)2×2 if aij = i – j is …………
Answer:
Question 19.
If A is of order 4 × 3 and B is of order 3 × 4 then the order of BA is ………………….
(1) 3 × 4
(2) 4 × 4
(3) 3 × 3
(4) 4 × 1
Answer:
(3) 3 × 3
Question 20.
If ……………..
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(4) 4
II. Answer the following.
Question 1.
Solve x + y = 7; y + z = 4; z + x = 1
Answer:
x + y = 7 ……(1)
y + z = 4 ………(2)
z + x = 1 …………(3)
Adding (1); (2) and (3)
2x + 2y + 2z = 12
x + y + z = 6 ….(4)
From (1) ⇒ x + y = 7
7 + z = 6
z = 6 – 7 = -1
From (2) ⇒ x + 4 = 6
x = 6 – 4 = 2
From (3) ⇒ y + 1 = 6
y = 6 – 1 = 5
The value of x = 2, y = 5 and z = -1
Question 2.
Find the HCF of 25x4y7; 35x3y8; 45x3y3
Answer:
25x4y7 = 5 × 5 × x4 × y7
35x3y8 = 5 × 7 × x3 × y8
45 x3y3 = 3 × 3 × 5 × x3 × y3
H.C.F. = 5x3y3
Question 3.
Find the values of k for which the following equation has equal roots.
(k – 12)x2 + 2(k – 12)x + 2 = 0
Solution:
(k−12)ax2+2(k−12)bx+2c=0
Δ = b2 – 4ac = (2(k – 12))2 – 4(6 – 12)(2)
= 4(k – 12)[(k – 12) – 2]
= 4(k – 12)(k – 14)
The given equation will have equal roots, if A = 0
⇒ 4(k – 12)(k – 14) = 0
k – 12 = 0 or k – 14 = 0
k = 12, 14
Question 4.
Find the LCM of x3 + y3; x3 – y3; x4 + x2y2 + y4
Answer:
x3 + y3 = (x + y) (x2 – xy + y2)
x3 – y3 = (x – y)(x2 + xy + y2)
x4 + x2y2 + y4 = (x2 + y2)2 – (xy)2
= (x2 + y2 + xy)
L.C.M. = (x + y)(x – y) (x2 + xy + y2)
(x2 – xy + y2)
= [(x + y) (x2 – xy +y2)]
[(x – y) (x2 + xy + y2)]
= (x3 + y3) (x3 – y3)
L.C.M. = x6 – y6
Question 5.
The sum of two numbers is 15. If the sum of their reciprocals is 310, find the numbers.
Solution:
Let the numbers be α, β
Sum of the roots = α + β = 15 ………….. (1)
sum of their reciprocals = 1α+1β=310 ……….. (2)
β+ααβ=310
10(α + β) = 3αβ …………. (3)
3αβ = 10 × 15 = 150
Products of the roots = αβ = 50 ………….. (4)
∴ From (1) & (4), we have
x2 – 15x + 50 = 0
(x – 10)(x – 5) = 0 ⇒ x = 10, 5
∴ he numbers are 10, 5.
Question 6.
For What value of k, the G.C.D. of [x2 + x – (2k + 2)] and 2x2 + kx – 12 is (x + 4)?
Answer:
p(x) = x2 + x – (2k + 2)
g(x) = 2x2 + kx – 12
G.C.D. = x + 4
when x + 4 is the G.C.D.
p(-4) = 0 or g(-4) = 0
[Hint: Take any one of the polynomial]
g(x) = 2x2 + kx – 12 = 0
2(-4)2 + k (-4) – 12 = 0
2(16) – 4x – 12 = 0
32 – 4k – 12 = 0
20 = 4k
k = 204 = 5
The value of k = 5
Question 7.
Simplify x2+x−6x2+4x+3
Answer:
x2 + x – 6 = (x + 3) (x – 2)
x2 + 4x + 3 = (x + 3) (x + 1)
Question 8.
Multiply
Answer:
if P = x3−36x2−49 and Q = x+6x+7 find the value of PQ.
Answer:
Question 10.
Simplify
xx+y – yx−y
Answer:
Question 11.
Find the square root of (x + 11)2 – 44x
Answer:
Question 12.
Find the square root of x4 + 1x4 + 2
Answer
Question 13.
Solve the equation 2x – 1 – 2x−2 = 3
Answer:
2x – 1 – 2x−2 = 3
(x – 3) (x – 1) = 0
x – 3 = 0 or x – 1 = 0
x = 3 or x = 1
The solution set is (1,3)
Question 14.
Find the roots of 2–√ x2 + 7x + 52–√ = 0
Answer:
2–√ x2 + 7x + 5 2–√ = 0
2–√ x2 + 2x + 5x + 5 2–√ = 0
2–√ x (x + 2–√) + 5 (x + 2–√) = 0
(x + 2–√) (2–√ x + 5) = 0
(x + 2–√ ) = 0 or 2–√ x + 5 = 0
x = – 2–√ or 2–√ x + 5 = 0
x = – 2–√ or 2–√ x = -5
x = −52√
The roots are and – 2–√ and −52√
Question 15.
Solve x+5−−−−−√ = 2x + 3 using formula method.
Answer:
x+5−−−−−√ = 2x + 3
(x+5−−−−−√)2 = (2x + 3)2
x + 5 = 4x2 + 9 + 12x
0 = 4x2 + 12x – x + 9 – 5
0 = 4x2 + 11x + 4
Here a = 4, b = 11, c = 4
Question 16.
The sum of a number and its reciprocal is 376. Find the number.
Answer:
Let the require number be “x”
Its reciprocal is 1x
By the given data
The required number is 16 or 6
Question 17.
Determine the nature of the roots of the equation 2x2 + x – 1 = 0
Answer:
2x2 + x – 1 = 0
Here a = 2,b = 1,c = -1
∆ = b2 – 4 ac
= 12 – 4(2) (-1)
= 1 + 8
= 9
Since b2 – 4ac > 0 the roots are real and unequal
Question 18.
Find the value of k for which the given equation 9x2 + 3kx + 4 = 0 has real and equal roots.
Answer:
9x2 + 3 kx + 4 = 0
a = 9, b = 5k, c = 4
since the equation has real and equal roots
b2 – 4ac = 0
(3k)2 – 4(9) (4) = 0
9k2 – 144 = 0
9k2 = 144
k2 = 1449 = 16
k = 16−−√
k = ± 4
Question 19.
If one root of the equation
3x2 – 10x + 3 = 0 is 13 find the other root
Answer:
α and β are the roots of the equation 3x2 – 10x + 3 = 0
Sum of the roots (α + β) = 103
Product of the roots (αβ) = 33 = 1
one of the roots is 13 (say α = 13)
αβ = 1
13 × β = 1
β = 3
The other roots is 3
Question 20.
Form the quadratic equation whose roots are 3 + 7–√; 3 – 7–√
Answer:
Sum of the roots = 3 + 7–√ + 3 – 7–√
= 6
Product of the roots = (3 + 7–√) (3 – 7–√ )
= 32 – (7–√)2
= 9 – 7
= 2
The required equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (6)x + 2 = 0
x – 6x + 2 = 0
Question 21.
If α and β are the roots of the equation 3x2 – 5x + 2 = 0, then find the value of α – β.
Answer:
α and β are the roots of the equation
3x2 – 5x + 2 = 0
α + β = 53, αβ = 23
Question 22.
Determine the matrix A = (aij)3×2 if aij = 3i – 2j
Answer:
aij = 3i – 2j
a11 = 3(1) – 2(1) = 3 – 2 = 1
a12 = 3(1) – 2(1) = 3 – 4 = 1
a21 = 3(2) – 2(1) = 6 – 2 = 4
a22 = 3(2) – 2(2) = 6 – 4 = 2
a31 = 3(3) – 2(1) = 9 – 2 = 7
a32 = 3(3) – 2(2) = 9 – 4 = 5
Question 23.
Answer:
Question 24.
Find if
Answer:
Question 25.
find BA
Answer:
Question 26.
Find the unknowns a, b, c, d, x, y in the given matrix equation.
Answer:
Equating the corresponding elements of the two matrices we get
d + 1 = 2
d = 2 – 1 = 1
10 + a = 2a + 1
10 – 1 = 2a – a
9 = a
36 – 2 = b – 5
3b – b = -5 + 2
2b = -3 ⇒ b = −32
a – 4 = 4c ⇒ a – 4c = 4
9 – 4c = 4 ⇒ 4c = 4 – 9
-4c = -5 ⇒ c = 54
The value of a = 9, b = −32, c = 54 and d = 1
Question 27.
Prove that
multiplication is inverse to each other.
Answer:
AB = BA = I
Multiplication of matrices are iverse to each other.
III. Answer the following questions.
Question 1.
Solve x – y5 = 6; y – z7 = 8; z – x2 = 10
Answer:
x – y5 = 6
multiply by 5
5x – y = 30 …….(1)
y – z7 = 8
Substitute the value of x = 8 in (1)
5(8) – y = 30
– y = 30 – 40 = -10
∴ y = 10
Substitute the value of x = 8 in (3)
2z – 8 = 20
2z = 20 + 8
z = 282 = 14
The value of x = 8, y = 10 and z = 14
Question 2.
Solve for x,y and z using the given 3 equations
2y – 4z + 3x = 3; 5x – 4y – 8z = 8 ; 6y + 6z +1x = 2
Answer:
Let 1x = a, 1y = b, 1z = c
3a + 2b – 4c = 3 ………(1)
5a – 4b – 8c = 8 ………(2)
a + 6b + 6c = 2 ………(3)
(1) × 2 ⇒ 6a + 4b – 8c = 6 …..(1)
Question 3.
100 pencils are to be kept inside three types of boxes A, B and C. If 5 boxes of type A, 3 boxes of type B, 2 boxes of type C are used 6 pencils are left out. If 3 boxes of type A, 5 boxes of type B, 2 boxes of type C are used 2 pencils are left out. If 2 boxes of type A, 4 boxes of type B and 4 boxes of type C are used, there is a space for 4 pencils. Find the number of pencils that each box can hold.
Answer:
Let the number of pencil in the box A be “x”
Let the number of pencil in the box B be “y”
Let the number of pencil in the box C be “z”
By the given first condition
5x + 3y + 2z = 94 ….(1)
By the given second condition
3x + 5y + 2z = 98 ….(2)
By the given third condition
2x + 4y + 4z = 104 ….(3)
subtract (1) and (3)
substitute x = 8 and y = 10 in (1)
5(8) + 3(10) + 2z = 94
40 + 30 + 2z = 94
2z = 94 – 70
2z = 24
z = 242 = 12
Number of pencil in box A = 8
Number of pencil in box B = 10
Number of pencil in box C = 12
Question 4.
What 2 masons earn in a day is earned by 3 male workers in a day. The daily wages of 15 female workers is ₹30 more than the total daily wages of 5 masons and 3 male workers. If one mason, one male worker and 2 female workers are engaged for a day, the builder has to pay ?160 as wages. Find the daily wages of a mason, a male worker and a female worker.
Answer:
Let the daily wage of a mason be ₹ x
Let the daily wage of a male worker be ₹ y
Let the daily wage of a female worker be ₹ z
By the given first condition
2x = 3y
2x – 3y = 0 …..(1)
By the given second condition
15z = 5x + 3y + 30
-5x – 3y + 15z = 30
5x + 3y – 15z = -30 ………(2)
By the given third condition
substitute the value of x = 60 in ….(1)
2(60) – 3y = 0
120 = 3y
y = 1203 = 40
substitute the value of x = 60 and y = 40 in (3)
60 + 40 + 2z = 160
2z = 160 – 100
2z = 60
z = 602 = 30
Daily wages of a manson = ₹60
Daily wages of a male worker = ₹40
Daily wages of a female worker = ₹30
Question 5.
Find the G.C.D. of x3 – 10x2 + 31x – 30 and 2x3 – 8x2 + 2x + 12
Answer:
p(x) = x3 – 10x2 + 31x – 30
g(x) = 2x3 – 8x2 + 2x + 12
= 2(x3 – 4x2 + x + 6)
G.C.D. = x2 – 5x + 6
Question 6.
The G.C.D of x4 + 3x3 + 5x2 + 26x + 56 and x4 + 2x3 – 4x2 – x + 28 is x2 + 5x + 7. Find their L.C.M.
Answer:
p(x) = x4 + 3x3 + 5x2 + 26x + 56
g(x) = x4 + 2x3 – 4x2 – x + 28
G.C.D. = x2 + 5x + 7
Question 7.
Find the values of “a” and “b” given that p(x) = (x2 + 3x + 2) (x2 – 4x + a); g(x) = (x2 – 6x + 9) × (x2 + 4x + b) and their G.C.D is (x + 2) (x – 3)
Answer:
p(x) = (x2 + 3x + 2) (x2 – 4x + a)
= (x + 1) (x + 2) (x2 – 4x + a)
G.C.D is given as (x + 2) (x – 3)
x – 3 is a factor of x2 – 4x + a
p(3) = 0
9 – 4(3) + a = 0
9 – 12 + a = 0
– 3 + a =0
a = 3
g(x) = (x2 – 6x + 9) (x2 + 4x + 6)
= (x – 3) (x – 3) (x2 + 4x + b)
But G.C.D. is (x + 2) (x – 3)
∴ x + 2 is a factor of x2 + 4x + 6
g(-2) = 0
4 + 4(-2) + b = 0
4 – 8 + 6 = 0
-4 + b = 0
b = 4
The value of a = 3 and b = 4
Question 8.
Find the other polynomial g(x), given that LCM, HCF and p(x) as (x – 1) (x – 2) (x2 – 3x + 3); x – 1 and x3 – 4x2 + 6x – 3 respectively.
Answer:
LC.M. = (x – 1) (x – 2) (x2 – 3x + 3)
HCF = (x – 1)
p(x) = x3 – 4x2 + 6x –
p(x) = (x – 1) (x2 – 3x + 3)
p(x) × g(x) = LCM × HCF
(x – 1) (x2 – 3x + 3) × g(x)
= (x – 1) (x – 2) (x2 – 3x + 3) (x – 1)
The other polynominal g(x) x2 – 3x + 2
Question 9.
Divide
Answer:
2x2 + x – 3 = 2x2 + 3x – 2x – 3
= x(2x + 3) – 1 (2x + 3)
= (2x + 3) (x – 1)
2x2 + 5x + 3 = 2x2 + 3x + 2x + 3
= x(2x + 3) + 1 (2x + 3)
= (2x + 3) (x + 1)
x2 -1 = (x + 1) (x – 1)
Question 10.
Simplify
Answer:
(x2 – x – 6) = (x – 3) (x + 2)
2x2 + 5x – 3 = 2×2 + 6x – x – 3
= 2x (x + 3) -1 (x + 3)
= (x + 3) (2x – 1)
x2 + 5x + 6 = (x + 2) (x + 3)
Question 11.
Find the square root of (6x2 + 5x – 6) (6x2 – x – 2) (4x2 + 8x + 3)
Answer:
6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) -2 (2x + 3)
= (2x + 3) (3x – 2)
6x2 – x – 2 = 6x2 – 4x + 3x – 2
= 2x (3x – 2) + 1 (3x – 2)
= (3x – 2) (2x + 1)
4x2 + 8x + 3 = 4x2 + 6x + 2x + 3
= 2x(2x + 3) + 1 (2x + 3)
= (2x + 3) (2x + 1)
Question 12.
Find the square root of the polynomial
4x2y2 + 8xy + 16 + 12 yx + 9y2x2
Answer:
Question 13.
If m – nx + 28x2 + 12x3 + 9x4 is a perfect square, then find the values of m and n.
Answer:
Arrange the polynomial in descending power of x.
9x4 + 12x3 + 28x2 – nx + m
Now,
Since the given polynomial is a perfect square,
-nx – 16x = 0
-x (n + 16) = 0
n + 16 = 0 ⇒ n = -16
m – 16 = 0 ⇒ m = 16
The value m = 16 and n = -16
Question 14.
If b + ax + 13x2 – 6x3 + 1x4 is a perfect square, find the values of “a” and “b”
Answer:
Arrange the values of “a” and “b”
Since it is a perfect square
ax + 12x = 0
1x (a + 12) = 0
a + 12 = 0 ⇒ a = -12
b – 4 = 0 ⇒ b = 4
The value of a = -12 and b = 4
Question 15.
Solve
1x+1 + 43x+6 = 23
Answer:
6x2 – 12x + 9x – 18 = 0
6x(x – 2) + 9(x – 2) = 0
(x – 2) (6x + 9) = 0
x – 2 = 0 or 6x + 9 = 0
x = 2 or 6x + 9 = 0
x = 2 or 6x = -9
x = – 96 = −32
The solution is −32 or 2
Question 16.
A two-digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number (solve by completing square method)
Answer:
Let the ten’s digit be “x”
∴ The unit digit = 14x
∴ The number is 10x + 14x
If the digits are interchanged the number is 140x + x
By the given condition
10x + 14x + 45 = 140x + x
multiply by x
10x2 + 14 + 45x = 140 + x2
9x2 + 45x + 14 – 140 = 0
9x2 + 45x – 126 = 0
Divided by 9
x2 + 5x – 14 = 0
x2 + 5x = 14
Since the digit of the number can not be negative
∴ x = 2
The number = 10x + 14x
= 20 + 142
= 20 + 7
= 27
The number is 27
Question 17.
A rectangular garden 10 m by 16 m is to be surrounded by a concreate walk of uniform width. Given that the area of walk is 120 sqm assuming the width of walk be ‘V form the equation then solve it by formula method.
Answer:
Area of the garden = 16 × 10
= 160 sq.m
Area of the garden with walking area
= (1.6 + 2x) (10 + 2x)
= 160 + 32x + 20x + 4x2
= 4x2 + 52x + 160
Area of the concrete walk = Area of the garden with walk – Area of garden
= 4x2 + 52x + 160 – 160
120 = 4x2 + 52x
4x2 + 52x – 120 = 0
(÷ by 4) ⇒ x2 + 13x – 30 = 0
Here a = 1, b = 13, c = -30
(comparing with ax2 + bx + c = 0)
Since the width cannot be negative. Width of the garden = 2 m
Question 18.
If α and β are the roots of the equation 3x2 – 5x + 2 = 0 find the value of
(i) αβ+βα
(ii) α – β
(iii) α2β+β2α
Answer:
Comparing with ax2 + bx + c = 0
a = 3, b = -5, c = 2
α and β are the roots of the equation
3x2 – 5x + 2 = 0
(iii) α2β+β2α=α3+β3αβ
Question 19.
If α and β are the roots of the equation 3x2 – 6x + 1 = 0 from the equation whose roots are
(i) α2 β;β2α
(ii) 2α + β; 2β + a
Answer:
α and β are the roots of 3x2 – 6x + 1 = 0
α + β = 63 = 2
αβ = 13
(i) Given the roots are α2β and β2α
Sum of the roots = α2β + β2α
= αβ (α + β)
= 13(2)
= 23
Product of the roots = (α2β) x (β2α)
= α2β2
= (αβ)3
= (13)3
= 127
The quadratic equation is
x2 – (sum of the roots) x + product of the roots = 0
x2 – (23) x + 127 = 0
multiply by 27
27x2 – 18x + 1 = 0
(ii) Given the roots are 2α + β; 2 β + α
Sum of the roots = 2α + β + 2 β + α
= 2(α + β) + (α + β)
= 2(2) + 2
= 6
The quadratic polynomial is
x2 – (sum of the roots) x + product of the roots = 0
x2 – 6x + 253 = 0
3x2 – 18x + 25 = 0
Question 20.
Find X and Y if
Answer:
Question 21.
Solve for x,y
Answer:
Question 22.
Show that A2 – 7A + 1013 = 0
Answer:
Question 23.
Verify that (AB)T = BT AT if
Answer:
From (1) and (2) we get
(AB)T = BTAT
Question 24.
Draw the graph of y = x2 and hence solve x2 – 4x – 5 = 0.
Answer:
Given equations are y = x2 and x2 – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.
(ii) Plot the points (- 4,16), (- 3, 9), (- 2,4), (-1, 1), (0,0), (1, 1), (2,4), (3, 9), (4,16), (5,25).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations
(v) The points of intersection of the line and the parabola are (-1, 1) and (5, 25).
The x-coordinates of the points are -1 and 5.
Thus solution set is {- 1, 5}.
Question 25.
Draw the graph of y = 2x2 + x – 6 and hence solve 2x2 + x – 10 = 0.
Answer:
Given equations are y = x2 and x2 – 4x – 5 = 0
(i) Assume the values of x from – 4 to 5.
(ii) Plot the points (- 4, 22), (- 3, 9), (- 2, 0), (-1, -5), (0, -6), (1, -3), (2, 4), (3, 15), (4, 30).
(iii) Join the points by a smooth curve.
(iv) Solve the given equations: Subtract 2x2 + x – 10 = 0
y = 4 is a straight line parallel to X-axis
(v) The straight line and parabola intersect at point (-2.5, 4) and (2, 4).
The x-coordinates of the points are -2.5 and 2.
The solution set is {- 2.5, 2}.
