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Question
Determine whether the given values of x is the solution of the given quadratic equation below:
6x2 - x - 2 = 0; x = `(2)/(3), -1`.
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Solution
6x2 - x - 2 = 0; x = `(2)/(3), -1`
Now put x = -1 in L.H.S. of equation.
L.H.S. = 6 x (-1)2 - (-1) -2
= 6 + 1 - 2
= 7 - 2 = 5 ≠ 0 ≠ R.H.S.
Hence, x = -1 is not a root of the equation.
Put x = `(2)/(3)` in L.H.S. of equation.
L.H.S. = 6 x `(2/3)^2 - (2)/(3) -2`
= `(24)/(9) - (2)/(3) - 2`
= `(8)/(3) - (2)/(3) - 2 = 0`
= 8 - 8 = 0
= R.H.S.
Hence, x = `(2)/(3)` is a solution of given equation.
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Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
