Quadratic equations Questions
1. 2x2-34x+144=0; 3y2-39y+108= 0 Which of the following cannot be value of x+y ?
A. 18
B. 13
C. 17
D. 14
Solution : 2x2-34x+144=0, dividing with 2 we get => x=9,8
3y2-39y+108=0, dividing with 3, we get => y=9,4
The possible values of x+y are 9+9=18, 9+8=17, 8+4=12, 9+4=13
The value of x+y=14 is not possible.
2. Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
A. -5, 3
B. 3, 5
C. -3, 5
D. -3, -5
E. 5, 2
Answer: Option A
x2 + 5x - 3x - 15 = 0
x(x + 5) - 3(x + 5) = 0
(x - 3)(x + 5) = 0
=> x = 3 or x = -5.
3. If the roots of a quadratic equation are 20 and -7, then find the equation?
A. x2 + 13x - 140 = 0
B. x2 - 13x + 140 = 0
C. x2 - 13x - 140 = 0
D. x2 + 13x + 140 = 0
Answer: Option C
Any quadratic equation is of the form
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is -140, the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
4. If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?
A. 3
B. 4
C. 5
D. 6
Answer: Option A
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = -(- 5/2) = 5/2 => a = 1/2
Product of the roots: 6a2 = b/2 => b = 12a2
a = 1/2, b = 3.
5. The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
A. 18, 20, 22
B. 20, 22, 24
C. 22, 24, 26
D. 24, 26, 28
Answer: Option B
Three consecutive even natural numbers be 2x - 2, 2x and 2x + 2.
(2x - 2)2 + (2x)2 + (2x + 2)2 = 1460
4x2 - 8x + 4 + 4x2 + 8x + 4 = 1460
12x2 = 1452 => x2 = 121 => x = ± 11
As the numbers are positive, 2x > 0. Hence x > 0. Hence x = 11.
Required numbers are 20, 22, 24.
6. 
A. -3, 1
B. 3, -1
C. 3, -2
D. 3, 2
Answer: Option A
7. A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
A. 10
B. 8
C. 15
D. 7.50
Answer: Option A
Let the price of each notebook be Rs.x.
Let the number of notebooks that can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x>0, x = 10.
8. I. a3 - 988 = 343,
II. b2 - 72 = 49 to solve both the equations to find the values of a and b?
A. If a > b
B. If a ≥ b
C. If a < b
D. If a ≤ b
E. If a = b or the relationship between a and b cannot be established.
Answer: Option B
a3 = 1331 => a = 11
b2 = 121 => b = ± 11
a ≥ b
9. 8m2+51m+18=0 ; 49n2-1=0
A. m<n
B. m<=n
C. m>n
D. m>=n
Solution : Option A
=> 8m2+51m+18=0
8m2 +48m+3m+18=0
8m(m +6) + 3(m +6)=0
(m+6) ( 8m+3)=0
m= -6 , -3/8
49n2-1=0
(7n-1) ( 7n+1)=0
n= 1/7, -1/7
Hence n>m
10. 4X2 + 24X + 36 =0 ; 5Y2 + 20Y + 20 =0
A. Y>X
B. x<=y
C. x>y
D. x>=y
Solution: Option A
4X2 + 24X + 36 =0
X2 + 6X + 9 =0
X2 + 3X + 3x + 9 =0
X = -3
5Y2 + 20Y + 20 =0
Y2 + 4Y + 4 =0
Y2 + 4Y + 4y + 4 =0
Y= -2
So, Y>X
11. I. x² − 16x + 64 = 0 ; y² − 16y + 63 = 0
A. y > x
B. x ≥ y
C. x > y
D. y ≥ x
Solution : Option C
=> (x − 2)² = 9
⇒ (x − 2) = ± 3
⇒ x = 5, −1
(2y + 8)² = 16
=>(2y + 8) = ± 4
⇒ y = −2, −6
x > y
12. (i) x² = 81 ; (ii) y² – 18y + 81 = 0
A. y>x
B. y>x
C. x>= y
D. y>= x
Solution : Option D
(i)x² = 81
x = ± 9
(ii)Y² – 18y + 81 = 0
(y – 9)² = 0
y = 9, 9
x ≤ y
13. Find the number which increased by 17 becomes equal to 60 times the reciprocal of the number.
Solution : x+17=60(1/x)
=> x(x+17)=60
X2-17x-60=0
X2+20x-3x-60
X(x+20)-3(x+20)=0
(x-3)(x+20)=0
X=3 or x=-20
14. If the roots of a quadratic equation are 20 and -7, then find the equation?
A. x2 + 13x - 140 = 0
B. x2 - 13x + 140 = 0
C. x2 - 13x - 140 = 0
D. x2 + 13x + 140 = 0
E. None of these
Solution: Any quadratic equation is of the form
x2 - (sum of the roots)x + (product of the roots) = 0 ---- (1)
sum of the roots is 13 and the product of the roots is -140,
the quadratic equation with roots as 20 and -7 is: x2 - 13x - 140 = 0.
15. A man could buy a certain number of notebooks for Rs.300. If each notebook cost is Rs.5 more, he could have bought 10 notebooks less for the same amount. Find the price of each notebook?
A. 10
B. 8
C. 15
D. 7.50
E. None of these
Solution: Let the price of each notebook be Rs.x.
Let the number of notebooks that can be brought for Rs.300 each at a price of Rs.x be y.
Hence xy = 300
=> y = 300/x
(x + 5)(y - 10) = 300 => xy + 5y - 10x - 50 = xy
=>5(300/x) - 10x - 50 = 0 => -150 + x2 + 5x = 0
multiplying both sides by -1/10x
=> x2 + 15x - 10x - 150 = 0
=> x(x + 15) - 10(x + 15) = 0
=> x = 10 or -15
As x > 0, x = 10.
16. x² – 34x + 288 = 0
y² – 28y + 192 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
Solution : C. X ≥ Y
x² – 34x + 288 = 0
x = 18, 16
y² – 28y + 192 = 0
y = 12, 16
17. x² – 31x + 234 = 0
y² – 34y + 285 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
Solution : E.
x² – 31x + 234 = 0
x = 13, 18
y² – 34y + 285 = 0
y = 15, 19
18. x² – 44x + 448 = 0
y² – 28y + 195 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
Answer . X > Y
x² – 44x + 448 = 0
x = 28, 16
y² – 28y + 195 = 0
y = 13, 15
19. x² – 31x + 234 = 0
y² – 34y + 285 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
Answer :X = Y or relation cannot be established
x² – 31x + 234 = 0
x = 13, 18
y² – 34y + 285 = 0
y = 15, 19
20. x² = 64
y² – 30y + 225 = 0
A. X > Y
B. X < Y
C. X ≥ Y
D. X ≤ Y
E. X = Y or relation cannot be established
Solution : B. X < Y
x² = 64
x = 8, – 8
y² – 30y + 225 = 0
y = 15, 15
21. In the question below, 2 equations are given. Solve both the equation and choose the correct option.
I. 4x4 – 5x2 + 1 = 0
II. 2y2 + 11y + 14 = 0
A. x > y
B. x ≥ y
C. x < y
D. x ≤ y
E. x = y or Cannot be determined.
Solution : (a)
4x4 – 5x2 + 1 = 0
(x2 – 1)(42 – 1) = 0
x = +1, -1, 1/2 or -1/2
2y2 + 11y + 14 = 0
(y + 2)(2y + 7) = 0
y = -2, -7/2
Least value of x = -1 > -2 = Largest value of y
So, x > y
22. I) 5x2 – 20x + 15 = 0 ; II) y + ∛0.125 = 0.9
Solution : I) 5x2 – 20x + 15 = 0
5x2 – 5x – 15x + 15 = 0
5x (x – 1) – 15 (x – 1) = 0
(5x – 15) (x – 1) = 0
x = 3, 1
II) y + ∛0.125 = 0.9
y + 0.5 = 0.9
y = 0.9 – 0.5 = 0.4
x > y
23. I) 5√2x2 + x - 3√2 = 0; II) 8y2 + 10y – 7 = 0
Solution : I) 5√2x2 + x - 3√2 = 0
5√2x2 – 5x + 6x - 3√2 = 0
5x (√2x – 1) + 3√2 (√2x – 1) = 0
(5x + 3√2) (√2x – 1) = 0
x = -3√2/5, 1/√2 = -0.84, 0.707
II) 8y2 + 10y – 7 = 0
8y2 - 4y + 14y – 7 = 0
4y (2y – 1) + 7(2y – 1) = 0
(4y + 7) (2y – 1) = 0
y = -7/4, ½ = -1.75, 0.5
Cant be determined
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