For a quadratic equation, a real number α is called the root of a quadratic equation ax^{2}+bx+c =0. Hence, we can write aα^{2} + bα + c = 0. So, x= α is the solution of a quadratic equation or the root of a quadratic equation. In other words, α satisfies the given quadratic equation.
Note: The zeros of the quadratic equation ax^{2}+bx+c = 0 are the same as the root of the quadratic equation ax^{2}+bx+c = 0.
Example: Solve the quadratic equation 2x^{2}+x300 = 0 by the factorisation method.
Sol:
Given quadratic equation: 2x^{2}+x300 = 0
By using factorisation, the quadratic equation 2x2+x 300 = 0 is written as:
2x^{2} – 24x+25x 300 = 0
2x(x12) +25(x12) =0
(i.e) (x12)(2x+25) = 0
Therefore, x12=0 and 2x+25 = 0
x12 = 0
Therefore, x= 12.
Similarly, 2x+25 = 0
2x= 25
x =25/2
x = 12.5.
Hence, the roots of the quadratic equation 2x2+x300 = 0 are 12 and 12.5.
Example: Find the roots of quadratic equation x2  7x + 10 = 0 using quadratic formula.
Sol:
Here, a = 1, b = 7 and c = 10. Then by quadratic formula:
x = [(7) ± √((7)2  4(1)(10))] / (2(1))
= [ 7 ± √(49  40) ] / 2
= [ 7 ± √(9) ] / 2
= [ 7 ± 3 ] / 2
= (7 + 3) / 2, (7  3) / 2
= 10/2, 4/2
= 5, 2
Therefore, x = 2, x = 5.
Example: The roots of the quadratic equation 3x210x+3 = 0 are
a) real and equal
b) imaginary
c) real, unequal and rational
d) none of these
Sol:
Given equation 3x^{2}10x+3 = 0
Here discriminant, D = b^{2}4ac
= (10)^{2} – 4×3×3
= 100 – 36
= 64
D is positive and a perfect square.
So the roots of the quadratic equation are real, unequal and rational.
Hence option (c) is the answer.
If α and β are the roots of the quadratic equation, then Quadratic equation is
x^{2} – (α + β) x + αβ = 0
OR
x^{2} – (sum of roots) x + product of roots = 0
where,
Example: If α and β are the roots of the equation x^{2}  4x + 2 = 0, find the value of
i) α^{2} + β^{2}
ii) α^{2}  β^{2}
iii) α^{3}  β^{3}
^{iv)1/α + 1/ β}
^{Sol:}
The given equation is x^{2}  4x + 2 = 0 ...................... (i)
According to the problem, α and β are the roots of the equation (i)
Therefore,
(i) Now α^{2} + β^{2} = (α + β)^{2}  2αβ = (4)^{2} – 2 x 2 = 16 – 4 = 12.
(ii) α^{2}  β^{2} = (α + β)( α  β)
Now (α  β)2 = (α + β)^{2}  4αβ = (4)^{2} – 4 x 2 = 16 – 8 = 8
⇒ α  β = ± √8
⇒ α  β = ± 2√2
Therefore, α^{2}  β^{2} = (α + β)( α  β) = 4 x (± 2√2) = ± 8√2.
(iii) α^{3} + β^{3} = (α + β)^{3}  3αβ(α + β) = (4)^{3} – 3 x 2 x 4 = 64 – 24 = 40.
(iv)
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