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Find the roots of the quadratic equation: x^{2} + 2x  15 = 0?
x^{2} + 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 (a^{2 }+ b^{2})x^{2 }− 2b(a + c)x + (b^{2}+c^{2}) = 0 are equal then
(a^{2 }+ b^{2})x^{2 }− 2b(a + c)x + (b^{2}+c^{2}) = 0
Roots are real and equal ∴ D = 0
D = b^{2 }− 4ac = 0
⇒ [−2b(a+c)]^{2 }− 4(a^{2 }+ b^{2})(b^{2 }+ c^{2}) = 0
⇒ b^{2}(a^{2 }+ c^{2 }+ 2ac) −(a^{2}b^{2} + a^{2}c^{2} + b^{4} + c^{2}c^{2}) = 0
⇒ b^{2}a^{2 }+ b^{2}c^{2 }+ 2acb^{2 }− a^{2}b^{2 }− a^{2}c^{2 }− b^{4} − b^{2}c^{2} = 0
⇒ 2acb^{2 }− a^{2}c^{2 }− 2acb^{2 }= 0
⇒ (b^{2 }− ac)^{2 }= 0
⇒ b^{2} = 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.
If the roots of a quadratic equation are 20 and 7, then find the equation?
Any quadratic equation is of the form: x^{2}  (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: x^{2}  13x  140 = 0.
The sum and the product of the roots of the quadratic equation x^{2} + 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 2x^{2}  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: 6a^{2} = b/2
⇒ b = 12a^{2} = 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.
⇒ x^{2} + (x + 1)^{2}  x(x + 1) = 91
⇒ x^{2} + 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 x^{2}  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: 3x^{2} = 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
⇒ 4x^{2}  8x + 4 + 4x^{2} + 8x + 4 = 1460
⇒ 12x^{2} = 1452
⇒ x^{2} = 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, x^{2 }+ 2ax + (10 − 3a) > 0, then the interval in which a lies, is?
In f(x) = ax^{2} + bx + c
When a > 0 and D < 0
Then f(x) is always positive.
x^{2} + 2ax + 10 − 3a > 0, ∀x ∈ R
⇒ D < 0
⇒ 4a^{2} − 4(10 − 3a) < 0
⇒ a^{2} + 3a − 10 < 0
⇒ (a+5)(a−2) < 0
⇒ a ∈ (−5,2)
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