The roots of the equation (x – a) (x – b) + (x – b) (x – c) + (x – c) (x – a) = 0 are :
Solution of a quadratic equation x^{2}+ 5x  6 = 0
If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of (p + q) terms will be
The sum of first p term of AP is q, which means that the number of terms is p
Thereby, let us take the first term as A and the common difference d
The sum of first q term of AP is p, which means that the number of terms is q
Thereby, let us take the first term as A and the common difference d
Therefore, the sum
Subtracting the sum of the term , p and q
After solving the equation we get the value of d as
Now with , first value of the series is a and the number of terms is
Therefore, the sum is
If α,β are the roots of the equation x^{2} + 2x + 4 = 0, then is equal to :
If then :
If α, β are the roots of the equation x^{2} + 7x + 12 = 0, then the equation whose roots are (α + β)^{2} and (α – β)^{2} is:
The value of k (k > 0) for which the equations x^{2} + kx + 64 = 0 and x^{2} – 8x + k = 0 both will have real roots is :
Let D₁ and D₂ be the discriminant of the given equations and these will have equal roots only if D₁, D₂ ≥ 0
Or, if D₁ = (k² 4 × 64) ≥ 0 and D₂ = (64  4k) ≥ 0
Or, if k² ≥ 256 and 4k ≤ 64
Or, if k ≥ 16 and k ≤ 16
Or, k = 16
Hence the positive value of k is 16.
If α,β are roots of the quadratic equation x^{2} + bx – c = 0, then the equation whose roots are b and c is
Solve for y : 9y^{4} – 29y^{2} + 20 = 0
Solve for x : x^{6} – 26x^{3} – 27 = 0
If 7th and 13th term of an A.P. are 34 and 64 respectively, then its 18th term is
Solve : – = 3
Solve for x : :
Solve x : :
Solve for x : – x + 2 = , x ε R :
Solve for x : 3^{x+2} + 3^{x} = 10
Solve for x : (x + 1) (x + 2) (x + 3) (x + 4) = 24 (x ε R) :
The sum of all the real roots of the equation x – 2^{2} + x – 2 – 2 = 0 is :
If a, b ε {1, 2, 3, 4}, then the number of quadratic equations of the form ax^{2} + bx + 1 = 0, having real roots is :
The number of real solutions of x – is :
Ifthen x is equal to :
The quadratic equation 3x^{2} + 2(a^{2} + 1) x + a^{2} – 3a + 2 = 0 possesses roots of opposite sign then a lies in:
The equation has :
The number of real solutions of the equation 2x^{2} – 5x + 2 = 0 is :
The number of real roots of the equation (x – 1)^{2} + (x – 2)^{2} + (x – 3)^{2} = 0 :
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