Quadratic Equations: Solved examples

Q1. Find the roots of the equation: x2+ 6x + 5 = 0
A. 5, -1
B. 5, 1
C. -5, -1
D. -5, 1
Sol: Option C
Solution: x2+ 6x + 5 = 0 ⇒ (x + 5)(x + 1) = 0 x + 5 = 0 or x + 1 = 0 x = - 5 or x = - 1
Q2. Find the roots of the equation: 3x2 – 3 = 8x
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A. -1/3, -1
B. 3, 1
C. -3, -1
D. -1/3, 3
Sol: Option D
Solution: 3x2 – 3 = 8x
3x 2 – 8x – 3 = 0 ⇒ 3x2 – 9x + x – 3 = 0
3x(x – 3) + 1(x – 3) ⇒ (3x + 1)(x – 3) = 0
3x + 1 = 0 or x – 3 = 0 ⇒ x = -1/3 or x = 3
Q3. one root of the quadratic equation 8x2 – 28x + z = 0 is six times the other, find the value of z.
A. 12
B. 10
C. -8
D. 4
Sol: Option A
Solution: Here in this equation a = 8, b = - 28 and c = z.
From the formula, sum of the roots = -28/8 = −7/2 (i)
If one root is α, the other root is 6α and the sum of roots will be 7α
7α =7/2 ⇒ α = 1/2. Other root will be ½ × 6 = 3
Now the product of the roots will be = c/a = z/8
Product of roots α × 6α = 6α2 ⇒ 6(1/2)2
⇒ Now 6/4 = z/8 ⇒ z = 12.
Q4. If 16x2 – 24x + m = 0 have equal roots, find the value of m.
A. -5
B. 9
C. 2
D. 11
Sol: Option B
Solution: 16x2 – 24x + m = 0
Since it has equal roots, discriminant should be equal to zero.
(- 24)2 – 4 × 16 × m = 0 ⇒ 576 = 64m
⇒ m = 576/64 = 9.
Q5. If p and q are the roots of the equation x2 + px + q = 0, then which value of p is not possible?
A. p = 1
B. p = 1 or 0 or -1/2
C. p = –2
D. p = –2 or 0
Sol: Option B
Solution: Since p and q are the roots of the equation x2 + px + q = 0,
p 2 + p2 + q = 0 and q2 + pq + q = 0
⇒ 2p 2 + q = 0 and q(q + p + 1) = 0
⇒ 2p 2 + q = 0 and q = 0 or q = – p – 1
when we take 2p 2+ q = 0 and q = 0, we get p = 0.
Or when we take 2p 2+ q = 0 and q = – p – 1, we get:
2p 2– p – 1 = 0, which gives us p =1 or p = -1/2
Hence, there can be three values for p i.e. p = 0 or 1 or -1/2
Q6. If x1 and x2 are the roots of the quadratic equation ax2 + bx + c = 0, find the equation whose roots are 1/x1 and 1/x2.
A. cx2 + bx + a =0
B. cx2 + bx + c =0
C. cx2 - bx + c = 0
D. cx2 - bx - a = 0
Sol: Option A
Solution: Sum of roots = x1 + x2 = -b/a
Product of roots = x1 x2 = c/a
Required sum of roots = (1/x1 + 1/ x2 ) = ( x1 + x2 / x1 x2) = -b/a  × a/c = -b/c
Product of roots = (1/x1 × 1/ x2 ) = a/c
Hence the required equation is  x2 + bx/c + a/c = 0
Therefore, cx2 + bx+ a = 0
Q7. The roots of the equation are integers x2/2 + bx + c = 0 if:
A. b2 – 2c > 0 2
B. b2 – 2c is square of an integer & b is an integer
C. b & c are integers
D. b & c are even integers
Sol: Option B
Solution: Take the LCM and make the equation as
x2 + 2bx + 2c = 0. Its general solution will be
Quadratic Equations
Cancelling 2 from the numerator and denominator, we get -b ±√(b2 – 2c
Now both of the roots have to be integers and that is only possible if b happens to be an integer and
b2 – 2c happens to be a perfect square.
Q8. Let α and β be the roots of the equation x2 + x + 1 = 0, then the equation whose roots are 1/α and 1/β is
A. x2 + x + 1 = 0
B. x2 – x + 1 = 0
C. x2 – x – 1 = 0
D. x2 + x – 1 = 0
Sol: Option A
Solution: For the given equation the sum of roots i.e. α+ β is –1/1 = - 1.
For the given equation the product of the roots i.e. αβ is 1. Now for the second equation, the roots are given to be 1/α and 1/β.
The sum of the roots for the equation to be calculated is 1/α + 1/β = (β + α)/αβ.
Both of these values are already calculated i.e. – 1/1 = - 1. The product of the roots for the second equation is 1/α × 1/β = 1/αβ, which can be calculated as 1/1 = 1.
The new equation is x2 – (sum of roots) x + product of roots = 0 i.e. x2 – (-1) x + 1 = 0 ⇒ x2 + x + 1 = 0.
Q9. Consider the equation of the form x2 + bx + c = 0, given b may or may not be equal to c. The number of such equations that have real roots and have coefficients b and c in the set {1, 2, 3, 4, 5, 6} is
A. 20
B. 18
C. 17
D. 19
Sol: Option D
Solution: In order to have real roots the value of (B2 – 4AC)1/2 ≥ 0
Given A as 1, what you need to calculate is for what values of (B, C) the value of B2≥ 4C.
Take B as 6, all the given six values C can take.
Now take B as 5, C can take all the 6 values again.
Take B as 4, C can take the values from 1 to 4 i.e. 4 values.
Take B as 3, C can take only 1 and 2. Take B as 2, C can take the value 1 only.
Taking B as 1, no value of C makes it possible.
Thus the total number of ways becomes:
6 + 6 + 4 + 2 + 1 = 19 values
Q10. a and b are the roots of x2 x – 3 = 0. Form the equation whose roots are (3a + 1) and (3b + 1)
A. x2 + 5x - 23 = 0
B. x2 - 5x + 23 = 0
C. x2 + 5x + 23 = 0
D. x2 - 5x - 23 = 0
Sol: Option D
Solution: Given a + b = -(-1) and ab = -3/1 = -3
Equation whose roots are (3a+1), (3b+1) is
x2– (sum of roots)x + (product of roots) = 0
i.e. x 2– (3a + 3b + 2)x + (9ab + 3a + 3b + 1) = 0,
x2 – 5x – 23 = 0.
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