Find the value of k for which x=3 is a solution of the quadratic equation (k+2)x²-kx+6=0 hence find the other root of the equation
Answers
Question:
Find the value of k for which x = 3 is a solution of the quadratic equation (k+2)x² - kx + 6 = 0 and hence find the other root of the equation.
Answer:
k = - 4
Another root : x = -1
Note:
• An equation of degree 2 is know as quadratic equation .
• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.
• The maximum number of roots of an equation will be equal to its degree.
• A quadratic equation has atmost two roots.
• The general form of a quadratic equation is given as , ax² + bx + c = 0 .
• The discriminant of the quadratic equation is given as , D = b² - 4ac .
• If D = 0 , then the quadratic equation would have real and equal roots .
• If D > 0 , then the quadratic equation would have real and distinct roots .
• If D < 0 , then the quadratic equation would have imaginary roots .
Solution:
The given quadratic equation is ;
(k+2)x² - kx + 6 = 0
According to the question ,
x = 3 is a solution of the given quadratic equation , thus x = 3 will satisfy the given equation.
Thus,
=> (k+2)x² - kx + 6 = 0
=> (k+2)•3² - k•3 + 6 = 0
=> (k+2)•9 - 3k + 6 = 0
=> 9k + 18 - 3k + 6 = 0
=> 6k + 24 = 0
=> 6k = -24
=> k = -24/6
=> k = - 4
Hence,
The required value of k is - 4 .
Now,
Substituting the the value of k in the given quadratic equation, it will be become ,
=> (k+2)x² - kx + 6 = 0
=> (-4+2)x² - (-4)x + 6 = 0
=> -2x² + 4x + 6 = 0
=> 2x² - 4x - 6 = 0
=> 2(x² - 2x - 3) = 0
=> x² - 2x - 3 = 0
=> x² - 3x + x - 3 = 0
=> x(x-3) + (x-3) = 0
=> (x-3)(x+1) = 0
=> x = -1 , 3
Hence,
It is clear that the another root of the given quadratic equation is - 1 .