Find the roots of the quadratic equation: x2 + 2x - 15 = 0?
x2 + 5x - 3x - 15 = 0
⇒ x(x + 5) - 3(x + 5) = 0
⇒ (x - 3)(x + 5) = 0
⇒ x = 3 or x = -5.
If the roots of the equation (a2 + b2)x2 − 2b(a + c)x + (b2+c2) = 0 are equal then
(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 roots of the equation 3x2 - 12x + 10 = 0 are?
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.
If the roots of a quadratic equation are 20 and -7, then find the equation?
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.
The sum and the product of the roots of the quadratic equation x2 + 20x + 3 = 0 are?
Sum of the roots and the product of the roots are -20 and 3 respectively.
If the roots of the equation 2x2 - 5x + b = 0 are in the ratio of 2:3, then find the value of b?
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.
The sum of the squares of two consecutive positive integers exceeds their product by 91. Find the integers?
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.
One root of the quadratic equation x2 - 12x + a = 0, is thrice the other. Find the value of a?
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.
The sum of the square of the three consecutive even natural numbers is 1460. Find the numbers?
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.
For all x, x2 + 2ax + (10 − 3a) > 0, then the interval in which a lies, is?
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)
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