Solve the equation 3x² - 14x + 8 = 0 when
1. x ∈ N
2. x ∈ Q
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Answers
Answered by
0
Solution:-
=> 3x²-14x+8=0
=> 3x²-12x-2x+8=0
=> 3x(x-4)-2(x-4)=0
=> (3x-2)(x-4)=0
=> x=2/3 and x=4
=> When x ∈ N
As 4 ∈N and 2/3 not belongs to N
Therefore
- The equation has 4 as it's root.
=> When x ∈ Q
As 4,2/3 ∈ Q
Therefore
- The equation has 4, 2/3 as it's roots.
Answered by
2
(i) x2 – 5x + 6 = 0; 2, -3
Let us substitute the given values in the expression and check,
When, x = 2
x2 – 5x + 6 = 0
(2)2 – 5(2) + 6 = 0
4 – 10 + 6 = 0
0 = 0
∴ x = 0
When, x = -3
x2 – 5x + 6 = 0
(-3)2 – 5(-3) + 6 = 0
9 + 15 + 6 = 0
30 = 0
∴ x ≠ 0
Hence, the value x = 2 is the root of the equation.
And value x = -3 is not a root of the equation.
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