Test: Quadratic Equations- 1 - UPSC

x² + 5x - 3x - 15 = 0
⇒ x(x + 5) - 3(x + 5) = 0
⇒ (x - 3)(x + 5) = 0
⇒ x = 3 or x = -5.

(a² + b²)x² − 2b(a + c)x + (b²+c²) = 0
Roots are real and equal ∴ D = 0
D = b² − 4ac = 0

⇒ [−2b(a+c)]² − 4(a² + b²)(b² + c²) = 0
⇒ b²(a² + c² + 2ac) −(a²b² + a²c² + b⁴ + c²c²) = 0
⇒ b²a² + b²c² + 2acb² − a²b² − a²c² − b⁴ − b²c² = 0
⇒ 2acb² − a²c² − 2acb² = 0
⇒ (b² − ac)² = 0
⇒ b² = ac

The discriminant of the quadratic equation is (-12)² - 4(3)(10) i.e., 24.
As this is positive but not a perfect square, the roots are irrational and unequal.

Any quadratic equation is of the form: x² - (sum of the roots)x + (product of the roots) = 0
where x is a real variable.

As the 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.

Sum of the roots and the product of the roots are -20 and 3 respectively.

The Correct option is A: 3

Let the roots of the equation 2a and 3a respectively.
Sum of Roots: 2a + 3a = 5a = -(- 5/2) = 5/2 
⇒ a = 1/2
Product of the roots: 6a² = b/2 
⇒ b = 12a² = 3
Hence, the values are: a = 1/2, b = 3.

Let the two consecutive positive integers be x and x + 1.

⇒ x² + (x + 1)² - x(x + 1) = 91
⇒ x² + x - 90 = 0
⇒ (x + 10)(x - 9) = 0
⇒ x = -10 or 9.
x = 9 [∵ x is positive]

Hence the two consecutive positive integers are 9 and 10.

Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12

⇒ x + 3x = 4x = 12
⇒ x = 3

Product of the roots: 3x² = 3(3)² = 27.

Let 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
⇒ x = 11 [∵ The numbers are positive, i.e. 2x > 0 ⇒ x > 0]

Thus, Required numbers are 20, 22, 24.

In f(x) = ax² + bx + c
When a > 0 and D < 0
Then f(x) is always positive.
x² + 2ax + 10 − 3a > 0, ∀x ∈ R

⇒ D < 0
⇒ 4a² − 4(10 − 3a) < 0
⇒ a² + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)