if one zero of polynomial x^2-12x + (3k-1) is 5 times the other then value of k is....
Answers
EXPLANATION.
One zero of the polynomial.
⇒ x² - 12x + (3k - 1) = 0. is 5 times others.
As we know that,
Let, one roots be = α.
Other roots be = 5α.
Sm of the zeroes of the quadratic polynomial.
⇒ α + β = - b/a.
⇒ α + 5α = - (-12/1).
⇒ α + 5α = 12.
⇒ 6α = 12.
⇒ α = 2.
Products of the zeroes of the quadratic polynomial.
⇒ αβ = c/a.
⇒ (α) x (5α) = (3k - 1)/1.
⇒ 5α² = 3k - 1.
Put the value of α = 2 in the equation, we get.
⇒ 5(2)² = 3k - 1.
⇒ 5 x 4 = 3k - 1.
⇒ 20 = 3k - 1.
⇒ 20 + 1 = 3k.
⇒ 21 = 3k.
⇒ k = 7.
Value of k = 7.
MORE INFORMATION.
Conjugate roots.
(1) = If D < 0.
One roots = α + iβ.
Other roots = α - iβ.
(2) = If D > 0.
One roots = α + √β.
Other roots = α - √β.
SOLUTION
GIVEN
One zero of the polynomial x² - 12x + (3k-1) is five times the other
TO DETERMINE
The value of k
EVALUATION
Here the given Quadratic polynomial is
x² - 12x + (3k-1)
Comparing with the general quadratic polynomial ax² + bx + c we get a = 1 , b = - 12 , c = 3k - 1
Now it is given that one zero of the polynomial is five times the other
Let the zeroes are α and 5α
Sum of the zeroes = - b/a
⇒ α + 5α = 12/1
⇒ 6α = 12
⇒ α = 2
So the zeroes are 2 and 10
Now
Product of the zeroes = c/a
⇒ 2 × 10 = 3k - 1
⇒ 3k - 1 = 20
⇒ 3k = 21
⇒ k = 7
FINAL ANSWER
Hence the required value of k = 7
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