Find he value of kfr which the quadrtic polynomial 9x2-3kx+k has has equal zeos
Answers
Question:
Find he value(values) of k for which the quadrtic polynomial 9x² - 3kx + k has has equal zeros.
Answer:
k = 0 , 4 .
Note:
• The values of x for which the polynomial p(x) becomes zero are called the zeros of the polynomial p(x).
• To find the zeros of the given polynomial p(x) , equate it to zero and then find the possible values of x , ie; operate on p(x) = 0.
• A quadratic polynomial can have at most two zeros .
• The discriminantnant D of a quadratic polynomial ,
p(x) = ax² + bx + c is given by;
D = b² - 4ac .
• If D = 0 , then both the zeros of the quadratic polynomial are real and equal.
• If D > 0 , then both the zeros of the quadratic polynomial are real and distinct.
• If D < 0 , then both the zeros of the quadratic polynomial are imaginary.
Solution:
The given quadratic polynomial is;
9x² - 3kx + k = p(x) { say }
Also,
The discriminant of the given quadratic polynomial will be;
=> D = (-3k)² - 4•9•k
=> D = 9k² - 9•4k
=> D = 9k(k - 4)
It is given that ,
The given quadratic polynomial has equal zeros , hence its discriminant must be zero.
ie ;
=> D = 0
=> 9k(k - 4) = 0
=> k = 0 OR (k - 4) = 0
=> k = 0 OR k = 4
Hence,
The required values of k are :
0 and 4 .
Moreover;
Case1 : When k = 0
Then , the given quadratic polynomial will reduce to ;
=> p(x) = 9x² - 3kx + k
=> p(x) = 9x² - 3•0•x + 0
=> p(x) = 9x²
Also,
For zeros of the polynomial;
=> p(x) = 0
=> 9x² = 0
=> x² = 0
=> x = 0,0
Case2 : When k = 4
Then , the given quadratic polynomial will reduce to ;
=> p(x) = 9x² - 3•4•x + 4
=> p(x) = 9x² - 12x + 4
=> p(x) = (3x)² - 2•(3x)•2 + 2²
=> p(x) = (3x - 2)²
Also,
For zeros of the polynomial;
=> p(x) = 0
=> (3x - 2)² = 0
=> (3x - 2)(3x - 2) = 0
=> x = 2/3 , 2/3
Hence,
If k = 0 , then the given polynomial will reduce to 9x² and it equal zeros will be 0,0 .
If k = 4 , then the given polynomial will reduce to (3x - 2)² and its equal zeros will be 2/3 , 2/3 .
Question :--- Find he value of k For which the quadrtic polynomial 9x²-3kx+k has equal zeroes ?
Concept used :----
If A•x^2 + B•x + C = 0 ,is any quadratic equation,
then its discriminant is given by;
D = B^2 - 4•A•C
• If D = 0 , then the given quadratic equation has real and equal roots.
• If D > 0 , then the given quadratic equation has real and distinct roots.
• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...
_____________________________
Solution :----
Our Equation is = 9x²-3kx+k
Comparing it with Ax^2 + Bx + C we get,
→ A = 9
→ B = (-3k)
→ C = k
since, Its Roots or zeros are Equal,
→ B² - 4AC = 0
Putting values now we get,
→ (-3k)² - 4*9*k = 0
→ 9k² - 36k = 0
→ 9k² = 36k
→ (9k)×k = 9k*4
Dividing both side by 9k now,
→ k = 4
Now,
→ (-3k)² - 4*9k = 0
→ 9k² - 4*9k = 0
Taking 9k common , we get,
→ 9k(k-4) = 0
If i we put 9k = 0,
we get, K = 0
Hence, we can say that, value of k will be 4 and 0 both ...