KSEEB Solutions for Class 10 Maths Chapter 10 Quadratic Equations Additional Questions
Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Additional Questions
I. Multiple Choice Questions:
Question 1.
The degree of a quadratic equations is
a. 1
b. 2
c. 3
d. 4
Answer:
b. 2
Question 2.
The standard form of quadratic equation is
a. ax2 + bx + c = 0
b. ax + c = 0
c. ax3 + bx2 + c = 0
d. ax4 + bx3 + cx2 + dx + c= 0
Answer:
a. ax2 + bx + c = 0
Question 3.
The standard form of pure quadratic equation is
a. ax2 + bx + c = 0
b. ax2 + c = 0
c. ax3 + bx2 + cx = 0
d. ax4 + bx3 + cx2 + dx + c= 0
Answer:
b. ax2 + c = 0
Question 4.
An quadratic equation has only
a. 1 root
b. 2 roots
c. 3 roots
d. 4 roots
Answer:
b. 2 roots
Question 5.
The roots of the quadratic equation 3x2 – 6x = 0 are
a. (0, 2)
b. (3, 6)
c. (0, – 2)
d. (0, 6)
Answer:
a. (0, 2)
Question 8.
The consecutive even integers are
a. (2x) (x + 2)
b. (x) (x + 2)
c. (x) (x + 1)
d. (x) (x – 1)
Answer:
b. (x) (x + 2)
Question 9.
Two consecutive positive integers differ by
a. 2
b. 1
c. 3
d. 4
Answer:
b. 1
Question 10.
The sum of the squares of two consecutive natural numbers is 25 represent this statement in the form of a quadratic equation.
a. x2 + (x + 1)2 = 25
b. x2 – (x – 1)2 = 25
c. (x + 1) – x2 = 25
d. x2 + (x + 1)2 + 25 = 0
Answer:
a. x2 + (x + 1)2 = 25
Question 11.
The discriminant of the quadratic equation ax2 + bx + c = 0 is
a. ∆ = b2 + 4ac
b. ∆ = b2 – 4ac
c. ∆ = 4abc
d. ∆ = b2 × 4ac
Answer:
b. ∆ = b2 – 4ac
Question 12.
The nature of the roots of the quadratic equation 2x2 – 4x + 3 = 0 are
a. real and distinct
b. real and equal
c. no real roots
d. imaginary
Answer:
d. imaginary
Question 15.
The quadratic equation whose roots are 2 and 3 is
a. x2 + 5x + 6 = 0
b. x2 – 5x + 6 = 0
c. x2 – 5x – 6 = 0
d. x2 + 5x – 6 = 0
Answer:
b. x2 – 5x + 6 = 0
Question 16.
The sum of the roots of the equation 2x2 – 8 = 0 is
a. 2
b. 4
c. – 4
d. 0
Answer:
d. 0
Question 17.
The product of the equation 3x2 = 9x is
a. 0
b. – 3
c. 2
d. 9
Answer:
a. 0
Question 19.
The name of the graph of y = 2x + 3 is called
a. Parabola
b. Hyper bola
c. Straight line
d. Bar graph
Answer:
c. Straight line
Question 20.
The name of the graph y = mx + c is called
a. Straight line
b. Parabola
c. Hyper bola
d. Pie chart
Answer:
a. Straight line
II. Short Answer Questions:
Question 1.
Write the standard form of Quadratic equation
Answer:
ax2 + bx + c = 0
Question 2.
If b2 – 4ac < 0, then find the nature of roots of Quadratic equation ax2 + bx + c = 0.
Answer:
Roots are imaginary
Question 5.
Find the discriminant of x2 + 5x + 5 = 0
Answer:
x2 + 5x + 5 = 0
It is in the form of ax2 + bx + c = 0
a = 1, b = 5 and c = 5
∆ = b2 – 4ac
= (5)2 – 4 (1) (5) = 25 – 20
∆ = 5
Discriminant = 5
Question 8.
Is x = – 2 is a solution of 3x2 + 13x + 14 = 0
Answer:
LHS 3x2 + 13x + 14
= 3 (- 2)2 + 13 (- 2) + 14
= 3 × 4 – 26 + 14
= 12 – 26 + 14
= 26 – 26 = 0
∴ x = – 2 is a solution
Question 9.
What is the condition for an equation of the form ax2 + bx + c = 0 to become a linear equation.
Answer:
a = 0
Question 10.
Write the formula to find the nature of roots?
Answer:
discriminat = b2 – 4ac
Question 11.
Write the formula to form a Quadratic equation if their roots m and n are given.
Answer:
x2 – (m + n) x + mn = 0
III. Long Answer Questions:
Question 4.
Solve 2x2 + 5x – 3 = 0 by completing squares method.
Answer:
Question 5.
Solve (2x + 3) (3x – 2) + 2 = 0 using formula.
Answer:
(2x + 3) (3x – 2) + 2 = 0
6x2 – 4x + 9x – 6 + 2 = 0
6x2 + 5x – 4 = 0
It is in the form of ax2 + bx + c = 0
a = 6, b = 5 and c = – 4
Question 6.
Solve a (x2 + 1) = x (a2 + 1) using formula.
Answer:
a(x2 + 1) = x (a2 + 1)
ax2 + a = x (a2) + 1
ax2 – (a2 + 1)x + a = 0
It is in the form of ax2 + bx + c = 0
a = a, b = – (a2 + 1) and c = a
Question 7.
For what Positive value of ‘m’ roots x2 – mx + 9 = 0 are
i. Equal
ii. distinct
iii. imaginary.
Answer:
x2 – mx + 9 = 0
It is in the form of ax2 + bx + c = 0
a = 1, b = – m and c = 9
∆ = b2 – 4ac
∆ = – (m)2 – 4 × 1 × 9
∆ = m2 – 36
i) If roots are equal
∆ = 0
m2 – 36 = 0
m2 = 36
m = √36 = ±6
m = 6
ii) If roots are distinct
∆ > 0
m2 – 36 > 0
m2 > 36
m > ±6
∴ m > 6
iii) If roots are imaginary
∆ < 0
m2 – 36 < 0
m2 < 36
∴ m < ±6
∴ m < 6
Question 8.
Answer:
10x2 + 12x + 36x – 120 – x2 = 0
9x2 + 36x – 108 = 0 divide by 9
x2 + 4x – 12 = 0
x2 + 6x – 2x – 12 = 0
(x + 6) – 2 (x + 6) = 0
(x + 6) (x – 2) = 0
x = – 6 (or) x = 2
Question 9.
Find the value of ‘P’ in a Quadratic equation (3P + 1) c2 + 2 (P + 1) c + P = 0 has equal roots.
Answer:
It is in the form of ax2 + bx + c = 0
a = (3P + 1) , b = 2(P + 1) and c = P
∆ = b2 – 4ac = 0 [has equal roots]
[2(P + 1)]2 – 4 × (3P + 1) P = 0
4(P2 + 2P + 1) – 4P(3P + 1) = 0
4[P2 + 2P + 1 – 3P2 – P] = 0
divide by 4
– 2P2 + P + 1 = 0 divide by – 1
2P2 – P – 1 = 0
2P2 – 2P + P – 1 = 0
2P(P – 1) + 1 (P – 1) = 0
(P – 1) (2P + 1) = 0
(P – 1) = 0 (or) 2P + 1 = 0
P = 1 (or) 2P = – 1
P = −1/2
∴ P = 1 (or) −1/2
Question 10.
In Rhombus ABCD the diagonals AC and BC intersect at E. If AE = x, BE = x + 7 and AB = x + 8. Find the lengths of the diagonals AC and BC.
Answer:
In Rhombus ABCD, AC and BD are diagonals are intersect at E.
x (x – 5) + 3 ( x – 5) = 0
(x – 5) (x + 3) = 0
x – 5 = 0 & x + 3 = 0
x = 5 & x = – 3
AE = x = 5 cm
∴ AC = 5 + 5 = 10 cm
BE = x + 7 = 5 + 7 = 12 cm
BD= 12 + 12 = 24 cm
∴ two diagonals are 10 cm and 24 cm.
Question 11.
A motor boat whose speed is 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes determine the speed of the stream.
Answer:
Let the speed of the stream be x km/hr
Speed of downstream = (15 + x) km/hr.
Time taken by the boat to go 30 km
downstream = 30/15+x hours
Speed upstream = 15 – x hours
Time taken by the boat to go 30 km
Upstream = 30/15–x hrs
It is given that the boat returns to the same point in 4 hours 30 minutes.
The speed of the stream = 5 km/hr.
Question 12.
A dealer sells an article for ₹ 24 and gains as much percent as the cost price of the article. Find the cost price of the article.
Answer:
Let the C.P of the article = x
S.P = ₹ 24
Gain = x% of C.P
x2 = 2400 – 100x
x2 + 100x – 2400 = 0
x2 + 120x – 20x – 2400 = 0
x(x + 120) – 20(x + 120) = 0
(x + 120) (x – 20) = 0
x + 120 = 0 & x – 20 = 0
x = – 120 & x = 20
∴ The C.P of the article = ₹ 20
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