How to find the zeroes and verify the relation of t square - 6t - 7
Answers
Question:
Find the zeros of the quadratic polynomial
t² - 6t - 7 and verify the relation with the coefficients.
Note:
∆ The general form of a quadratic polynomial is given as ; p(x) = ax² + bx + c .
∆ Zeros of a polynomial p(x) are the possible values of x for which the p(x) become zero.
∆ To find the zeros of polynomial p(x) , operate on p(x) = 0.
∆ The maximum number of zeros of a polynomial is equal to its degree.
∆ A quadratic polynomial will have at most two zero , as its degree is 2 .
∆ If A and B are the zeros of the quadratic polynomial p(x) = ax² + bx + c , then ;
• Sum of zeros,(A+B) = - b/a
• Product of zeros,(A•B) = c/a
∆ If A and B are given zeros of a quadratic polynomial p(x)., then p(x) will be given as ;
p(x) = x² - (A+B)x + A•B
Solution:
Here,
The given quadratic polynomial is ;
t² - 6t - 7.
Clearly ,
Coefficient of t² = 1 (ie, a = 1)
Coefficient of t = -6 (ie, b = -6)
Constant term = -7 (ie, c = -7)
Now,
In order to find the zeros of the given quadratic polynomial, equate it to zero .
Thus,
=> t² - 6t - 7 = 0
=> t² - 7t + t - 7 = 0
=> t(t-7) + (t-7) = 0
=> (t-7)(t+1) = 0
=> t = 7 , - 1
Hence,
The two zeros of the given quadratic polynomial are 7 and -1 .
Verification of the relation between the sum of zeros and coefficient:
Sum of zeros = 7 + (-1) = 7 - 1 = 6
Also, - b/a = -(-6)/1 = 6
Clearly,
Sum of zeros = -b/a
Verification of the relation between the product of zeros and coefficient:
Product of zeros = 7×(-1) = -7
Also, c/a = -7/1 = -7
Clearly,
Product of zeros = c/a
Hence verified.