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The solution of the cubic equation x3-6x2+11x-6=0 is given by the triplet:
- a)(-1, 1, -2)
- b)(-2, 2, 3)
- c)(1, 2, 3)
- d)(0, 4, -5)
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
The solution of the cubic equation x3-6x2+11x-6=0 is given by the trip...
x^3–6x^2+11x-6
=x^3-x^2–5x^2+5x+6x-6
=x^2(x-1)-5x(x-1)+6(x-1)
=(x-1)(x^2–5x+6)
=(x-1)(x^2–2x-3x+6)
=(x-1){x(x-2)-3(x-2)}
=(x-1){(x-2)(x-3)}
=(x-1)(x-2)(x-3)
=x^3-x^2–5x^2+5x+6x-6
=x^2(x-1)-5x(x-1)+6(x-1)
=(x-1)(x^2–5x+6)
=(x-1)(x^2–2x-3x+6)
=(x-1){x(x-2)-3(x-2)}
=(x-1){(x-2)(x-3)}
=(x-1)(x-2)(x-3)
Final solutions are
x=1, x=2, x=3
Or
x1=1,x2=2,x3=3
Most Upvoted Answer
The solution of the cubic equation x3-6x2+11x-6=0 is given by the trip...
Solution:
The given cubic equation is: x3-6x2+11x-6=0
To find the solution of this cubic equation, we can use the Rational Root Theorem.
Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q (where p and q are integers and q ≠ 0), then p must be a factor of the constant term and q must be a factor of the leading coefficient.
In this case, the constant term is -6 and the leading coefficient is 1. Therefore, the possible rational roots are:
±1, ±2, ±3, ±6 and their reciprocals.
We can try these roots one by one until we find a root that satisfies the equation.
Let's try x=1:
(1)3-6(1)2+11(1)-6 = 1-6+11-6 = 0
Therefore, x=1 is a root of the equation.
We can now use long division or synthetic division to factorize the cubic equation.
Using long division, we get:
x3-6x2+11x-6 = (x-1)(x2-5x+6)
We can now solve the quadratic equation x2-5x+6=0 using the quadratic formula:
x = (-(-5) ± √((-5)2-4(1)(6))) / (2(1))
x = (5 ± √1) / 2
x1 = 3, x2 = 2
Therefore, the solutions of the cubic equation x3-6x2+11x-6=0 are:
x1 = 1, x2 = 2, x3 = 3
Hence, the correct answer is option 'C'.
The given cubic equation is: x3-6x2+11x-6=0
To find the solution of this cubic equation, we can use the Rational Root Theorem.
Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root p/q (where p and q are integers and q ≠ 0), then p must be a factor of the constant term and q must be a factor of the leading coefficient.
In this case, the constant term is -6 and the leading coefficient is 1. Therefore, the possible rational roots are:
±1, ±2, ±3, ±6 and their reciprocals.
We can try these roots one by one until we find a root that satisfies the equation.
Let's try x=1:
(1)3-6(1)2+11(1)-6 = 1-6+11-6 = 0
Therefore, x=1 is a root of the equation.
We can now use long division or synthetic division to factorize the cubic equation.
Using long division, we get:
x3-6x2+11x-6 = (x-1)(x2-5x+6)
We can now solve the quadratic equation x2-5x+6=0 using the quadratic formula:
x = (-(-5) ± √((-5)2-4(1)(6))) / (2(1))
x = (5 ± √1) / 2
x1 = 3, x2 = 2
Therefore, the solutions of the cubic equation x3-6x2+11x-6=0 are:
x1 = 1, x2 = 2, x3 = 3
Hence, the correct answer is option 'C'.
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The solution of the cubic equation x3-6x2+11x-6=0 is given by the triplet:a)(-1, 1, -2)b)(-2, 2, 3)c)(1, 2, 3) d)(0, 4, -5)Correct answer is option 'C'. Can you explain this answer?
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