Please solve this question.
Answers
Step-by-step explanation:
Given :-
One of the root is 6 times the other root of the quadratic equation kx²-14x+8 = 0
To find :-
Find the value of k ?
Solution :-
Given quadratic equation is kx²-14x+8 = 0
On comparing this with the standard quadratic equation ax²+bx+c = 0
We have,
a = k
b = -14
c = 8
Let the other root be A
Then the one of the roots = 6 times the other
= 6A
We know that
Sum of the roots = -b/a
=> 6A+A = -(-14)/k
=> 7A = 14/k
=> A = (14/k)/7
=> A = 14/7k
=>A = 2/k -------------(1)
Product of the roots = c/a
=> 6A×A = 8/k
=> 6A² = 8/k ---------(2)
On Substituting the value of A from (1) in (2)
=> 6(2/k)² = 8/k
=> 6(4/k²) = 8/k
=>(6×4)/k² = 8/k
=> 24/k² = 8/k
On applying cross multiplication then
=> 8×k² = 24×k
=> k²/k = 24/8
=> (k×k)/k = (3×8)/8
=>k = 3
Therefore, k = 3
Answer :-
The value of k for the given problem is 3
Check:-
If k = 3 then the Quadratic equation is
3x²-14x+8 = 0
=> 3x² -12x-2x +8 = 0
=> 3x(x-4)-2(x-4) = 0
=> (x-4)(3x-2) = 0
=> x-4 = 0 or 3x-2 = 0
=> x = 4 or x = 2/3
The roots are 4 and 2/3
6 times 2/3
=> 6×2/3
=> 12/3
=> 4
One root is 6 times the other root
Verified the given relations in the given problem.
Used formulae:-
- The standard quadratic equation is ax²+bx+c = 0
- Sum of the roots = -b/a
- Product of the roots = c/a