Quadratic Equations For Class 10 Maths Important Questions
Q 1 – The quadratic equation x2 + 3x + 22 = 0 has
(a) two distinct real roots
(b) two equal real roots
(c) no real roots
(d) more than 2 real roots
(c) no real roots
Q 2 – Which of the following equations has 2 as a root?
(a) x2 – 4x + 5 = 0
(b) x2 + 3x – 12 = 0
(c) 2x2– 7x + 6 = 0
(d) 3x2– 6x – 2 = 0
(c) 2x2– 7x + 6 = 0
Q 3 – The real roots of the equation x 2/3 + x 1/3 – 2 = 0 are
(a) 1, 8
(b) – 1, – 8
(c) – 1, 8
(d) 1, – 8
(d) 1, – 8
Q 4 – Check whether the following are quadratic equations :
(i) x3 – 4x2 – x + 1 = (x – 2) 3
(ii) (2x – 1)(x – 3) = (x + 5)(x – 1)
(i) x3 − 4x2−x+1
= x3+ 3x2(−2)+3x(−2)2+(−2)3
x3−4x2−x+1=x3−6x2+12x−8
x3 − 4x2 − x + 1 − x3 + 6x2 −12x + 8 = 0
2x2−13x + 9 = 0
(ii) (x−3) (2x − 1)= (x − 1) (x + 5)
2x2−7x + 3 = x2 + 4x – 5
x2 − 11x + 8 = 0
Q 5 – Assertion: 4 x2 – 12x + 9 = 0 has repeated roots.
Reason: The quadratic equation ax2 + + bx + c = 0 have repeated roots if discriminant D > 0.
(a) Both assertion (A) and reason (R) are true and reason
(R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason
(R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(b) Both Assertion and Reason are correct and Reason is not the correct explanation for Assertion.
Q 6 – Find the nature of roots of the quadratic equation:
2x2 – 3√ 2 x + 9/4 = 0
Q 7 – Find the roots of the quadratic equation 3x2 – 2√6 x + 2 = 0.
Q 8 – If the quadratic equation px2 – 2px + 15 = 0 has two equal roots, then find the value of p.
Equation has equal roots (px²-2px + 15 = 0)
∴ b²- 4ac = 0
Comparing the equation with ax²+bx+c=0
a = p, b = -2√5p & c = 15
(-2√5p)² – 4(p) (15)=0
20p²- 60p = 0
20p ( p – 3 ) = 0
∴ p = 0,
&
P=3.
Q 9 – Find the discriminant of the quadratic equation 4 √2x2 + 8x + 2√2 = 0) .
D=b2− 4ac = (8)2 – 4 (4√2) (2√2)= 64 – 64 = 0
Q 10 – Show that x = – 2 is a solution of 3x2 + 13x + 14 = 0.
Substitute the value of x in equation,
3x2 + 13x + 14 = 0
3 (- 2) 2 + 13 x (- 2) + 14 = 0
3 x 4 – 26 + 14 = 0
– 26 + 26 = 0
Q 11 – Find the roots of the following quadratic equations by factorisation:
(i) x√2( x+√2) +5( x+√2)= 0
(x+√2)(x√2+5)=0
which can be reduced as
x= -√2, -5/√2
(ii) 16x2 − 8x + 1 = 0
=0
x = 1/4 and x = 1/4
Q 12 – The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let x be the base of the triangle, then the altitude will be (x−7).
By Pythagoras theorem,
x2+(x−7)2=(13)2
2x2−14x + 49 −169 = 0
2x2−14x−120 = 0
x2−7x – 60 = 0
x2−12x + 5x – 60 = 0
(x−12) (x+5)=0
x=12 ,x = − 5
Since the side of the triangle cannot be negative, so the base of the triangle is 12cm and the altitude of the triangle will be 12−7=5cm.
Q 13 – Solve for x: 
√3x2−2√2x − 2√3 = 0
√3x2−3√2x + 2√2x − 2√3 = 0
√3x(x−√6)+ √2(x−√6) = 0
either (√3x+√2) = 0 or (x−√6) = 0
x=√6 or −2√2/√3
Q 14 – If 2 is a root of the quadratic equation 3x2 + px – 8 = 0 and the quadratic equation 4x2 –2px + k = 0 has equal roots, find k.
Correct option is A)
+ px – 8 = 0
Substitute , since is root of equation
−2(−2) x + k = 0 [ Since, ]
+ 4x +k= 0
Here,
It is given that roots are equal.
− 4ac = 0
4(4) (k)=0
Q 15 – If ad ≠ bc, then prove that the equation (a2 + b2) x2 + 2(ac + bd) x + (c2 + d2) = 0 has no real roots.
(a2+b2)x2+2 (ac+ bd)x +c2+d2=0
4(a2+b2)(c2+d2)
c2 + b2d2 + 2acbd) − 4(a2c2 + b2d2+ a2d2+b2c2)
d2+b2c2−2acbd)
it has no real roots.
Q 16 – Is the following situation possible? If so, determine their present ages. The sum of the ages of the two friends is 20 years. Four years ago, the product of their age in years was 48.
Let their present ages be and yrs respectively.
Four years ago, their ages would be (x-4) and (16-x) yrs respectively.
(x−4) (16−x) = 48
16x − x2−64+ 4x = 48
x2− 20x + 112 = 0
b2− 4ac = − 48
Since the discriminant the roots would be imaginary.
Hence, such a situation is not possible.
Q 17 – Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the square.
Q 18 – A two-digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
Let the unit digit and tens digit of the two digit number be x and y respectively.
Number = 10y + x
According to the Question,
⇒ 10y + x = 4(y + x)
⇒ 10y + x = 4y + 4x
⇒ 10y – 4y = 4x – x
⇒ 6y = 3x
⇒ 2y = x … (i)
Also, 10y + x = 3xy …. (ii)
⇒ 10y + 2y = 3(2y)y [From Eq (i)]
⇒ 12y = 6y²
⇒ 6y² – 12y = 0
⇒ 6y(y – 2) = 0
⇒ y = 0, 2
Rejecting y = 0 as tens digit should not be zero for two digit number
⇒ x = 4
Number = 10y + x = 10 × 2 + 4 = 24.
Hence, the required number is 24.
Q 19 – A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig.). Find its length and breadth.
Let breadth of a rectangular Park= x m
Length of a rectangular Park=(x+3) m
Area of rectangle= Length × Breadth
Area of rectangular Park= Length × Breadth
= x(x+3)= (x²+3x)
Area of rectangular Park= (x²+3x) m²
Base of a rectangular Park= Breadth of the rectangular Park (Given)
Base of a rectangular Park= x m
Altitude of Triangular Park= 12 m (GIVEN)
Area of triangle= ½× base× altitude
Area of Triangular Park= 1/2× x×12= 6x m²
Area of Triangular Park= 6x m²
ATQ
Area of rectangular Park= 4 + area Triangular Park
x²+3x= 4+ 6x
x²+3x-6x-4=0
x²-3x-4=0
x²-4x+x -4=0
[By middle term splitting]
x(x-4)+1(x-4)=0
(x-4)(x+1)=0
x=4, x= -1
Breadth cannot be negative ,so neglect x= – 1
Breadth (x)=4
Hence, breadth of a rectangular Park= 4 m and length of the rectangular Park= x+3= 4+3= 7 m.
Q 20 – Find the roots of the following equation:
⇒ or
or
Q 21 – Find that non-zero value of k, for which the quadratic equation kx2 + 1 – 2(k – 1)x + x2 = 0 has equal roots. Hence find the roots of the equation.
. (k + 1)x2-2(k-1)x + 1 = 0
This is a quadratic equation of form ax2 + bx + c = 0
Where a = k + 1, b = -2(k-1), c = 1
For equal roots, D = 0
D = b2-4ac = 0
Substituting the values of a, b and c
(-2(k-1))2-4(k + 1)(1) = 0
⇒ 4(k2 + 1-2k)-4k-4 = 0
⇒ 4k2-12k = 0
⇒ 4k(k-3) = 0
There are two possible values of k, k = 0 and k = 3
We neglect k = 0 as on putting k = 0; the equation does not remain quadratic.
∴ k = 3 is our required value of k.
Now,
As we know roots are equal,
Therefore, each root =
Q 22 – Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5km/hr faster than the second train. If after two hours, they are 50 km apart, find the average speed of each train.
Let s = the speed of the northbound train
Then
(s+5) = the speed of the westbound train
:
This is a right triangle problem: a2 + b2 = c2
The distance between the trains is the hypotenuse
dist = speed * time
The time is 2 hrs, so we have
a = 2s; northbound train distance
b = 2(s+5) = (2s+10); westbound distance
c = 50; distance between the two trains
:
(2s) 2 + (2s+10)2 = 502
4s2 + 4s2 + 40s + 100 = 2500
Arrange as a quadratic equation4s2 + 4s2 + 40s + 100 – 2500 = 0
8s2 + 40s – 2400 = 0
:
Simplify, divide by 8:
s2 + 5s – 300 = 0
:
Factors to
(s – 15) (s + 20) = 0
:
The positive solution is what we want here
s = 15 mph is the speed of the northbound train
then
5 + 15 = 20 mph is the speed of the westbound train
:
:
Check this; find the distance (d) between the trains using these distances
Northbound traveled 2(15) = 30 mi
Westbound traveled 2(20) = 40 mi
d =
d = 50
Q 23 – If one root of the equation ax2 + bx + c = 0 is three times the other, then b2 : ac =
(a) 3 : 1
(b) 3 : 16
(c) 16 : 3
(d) 16 : 1
(c) 16 : 3
Q 24 – Solve the following quadratic equation for x : 
Q 25 – Solve for x :
Q 26 – Solve for x :
{2x/(x – 5) }² + 10x/(x – 5) – 24 = 0
multiplying by (x – 5)² both sides
=> 4x² + 10x(x – 5) – 24(x – 5)² = 0
Dividing by 2 both sides
=> 2x² + 5x(x – 5) – 12(x² -10x + 25) = 0
=> 2x² + 5x² – 25x – 12x² + 120x – 300 =0
=> -5x² +95x – 300 =0
=> x² – 19x + 60 = 0
=> x² -4x – 15x + 60 = 0
=> x(x – 4) – 15(x – 4) =0
=> (x – 15)(x – 4) =0
=> x = 4 or 15
Q 27 – Check whether the following are quadratic equations :
(i) x3 – 4x2 – x + 1 = (x – 2) 2
(ii) (2x – 1)(x – 3) = (x + 5)(x – 1)
Q 28 – A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Q 29 – Find the roots of the quadratic equation 3x2 – 2 x + 2 = 0.
Q 30 – If the quadratic equation px2 – 2px + 15 = 0 has two equal roots, then find the value of p.