Aptitude - Quadratic Equations
6.
If x2 − 3x + 1 = 0, then the value of x + 1x is :
Answer
x2 − 3x + 1 = 0 ⇒ x2 + 1 = 3x
⇒ x2 + 1x = 3
⇒ x + 1x = 3
Answer :
Option DExplanation :
x2 − 3x + 1 = 0 ⇒ x2 + 1 = 3x
⇒ x2 + 1x = 3
⇒ x + 1x = 3
7.
The roots of 2x2 − 6x + 3 = 0 are :
Answer
D = [(−6)2 − 4 x 2 x 3] = (36 − 24) = 12
Thus, D > 0 and not a perfect square.
∴ Roots are real, unequal and irrational.
Answer :
Option CExplanation :
D = [(−6)2 − 4 x 2 x 3] = (36 − 24) = 12
Thus, D > 0 and not a perfect square.
∴ Roots are real, unequal and irrational.
8.
If 2 + i √3 is a root of the equation x2 + px + q = 0, where p and q are real, then (p, q) is :
Answer
Let α = 2 + i √3. Then β = 2 − i √3
∴ α + β = 4, αβ = (2 + i √3)(2 − i √3) = (4 + 3) = 7
∴ Equation is x2 − 4x + 7 = 0
∴ p = −4, q = 7
Answer :
Option CExplanation :
Let α = 2 + i √3. Then β = 2 − i √3
∴ α + β = 4, αβ = (2 + i √3)(2 − i √3) = (4 + 3) = 7
∴ Equation is x2 − 4x + 7 = 0
∴ p = −4, q = 7
9.
If a, b are the two roots of a quadratic equation such that a + b = 24 and a − b = 8, then the quadratic equation is :
Answer
On solving a + b = 24 and a − b = 8, we get a = 16, b = 8.
∴ ab = 128
∴ Required equation is x2 − (a + b)x + ab = 0 i.e. x2 − 24x + 128 = 0
Answer :
Option BExplanation :
On solving a + b = 24 and a − b = 8, we get a = 16, b = 8.
∴ ab = 128
∴ Required equation is x2 − (a + b)x + ab = 0 i.e. x2 − 24x + 128 = 0
10.
If the roots of 2x2 + 3x + p = 0 be equal, then the value of p is :
Answer
D = (32 − 4 x 2 x p) = 9 − 8p
Since, roots are equal D = 0 ⇒ 9 − 8p = 0
∴ p = 98
Answer :
Option AExplanation :
D = (32 − 4 x 2 x p) = 9 − 8p
Since, roots are equal D = 0 ⇒ 9 − 8p = 0
∴ p = 98
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