what is the value of k in equation 8x² - 6x + k = 0 if one of its root will be square of other ?
Answers
Step-by-step explanation:
Given :-
The equation 8x^2 - 6x + k = 0 if one of its root will be square of other .
To find :-
What is the value of k ?
Solution :-
Given quardratic equation is 8x^2 - 6x + k = 0
Let P(x) = 8x^2 - 6x + k = 0
On Comparing this with the standard quadratic equation ax^2 +bx+c = 0
a = 8
b = -6
c = k
Let the other root of P(x) be α
Then the one of the roots = square of the other
= α^2
The roots of the given P(x) = α^2 and α
We know that
Sum of the roots = -b/a
α + α^2 = -(-6)/8
=> α + α^2 = 6/8
=>α + α^2 = 3/4
=> 4( α + α^2 ) = 3
=> 4α + 4α^2= 3
=> 4α^2 + 4α - 3 = 0
=> 4 α^2 + 6α - 2α - 3 = 0
=> 2 α( 2 α + 3) -1 ( 2 α +3 ) = 0
=> (2 α + 3)(2 α - 1) = 0
=> (2 α + 3) = 0 or (2 α - 1) = 0
=> 2 α = - 3 or 2 α = 1
=> α = -3/2 or α = /12-------(1)
And
Product of the roots = c/a
=>α × α^2 = k/8
=> α^3 = k/8 ------(2)
Put α = -3/2 in the equation (2) then
=> (-3/2)^3 = k/8
=> -27/8 = k/8
=> -27 = k
Therefore, k = -27
Put α = 1/2 in the equation (2) then
=> (1/2)^3 = k/8
=> 1/8 = k/8
=> 1 = k
Therefore, k = 1
The values of k = 1 and -27
Answer:-
The values of k for the given problem are 1 and -27
Used formulae:-
- The standard quadratic equation is
ax^2 +bx+c = 0
- Sum of the roots = -b/a
- Product of the roots = c/a
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