find the relationship is the quadratic polynomial 6x²-3-7x and verify the relationship between their 0 and coefficients About the author Eden
Answer: Given f(x) = 6×2 – 7x – 3 To find the zeros Let us put f(x) = 0 ⇒ 6×2 – 7x – 3 = 0 ⇒ 6×2 – 9x + 2x – 3 = 0 ⇒ 3x(2x – 3) + 1(2x – 3) = 0 ⇒ (2x – 3)(3x + 1) = 0 ⇒ 2x – 3 = 0 x = 3/2 ⇒ 3x + 1 = 0 ⇒ x = -1/3 It gives us 2 zeros, for x = 3/2 and x = -1/3 Hence, the zeros of the quadratic equation are 3/2 and -1/3. Now, for verification Sum of zeros = – coefficient of x / coefficient of x2 3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6 Product of roots = constant / coefficient of x2 3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2 Therefore, the relationship between zeros and their coefficients is verified. Reply
Answer:
695689688692810-%5863986/ -6836-6%* 648
Answer:
Given
f(x) = 6×2 – 7x – 3
To find the zeros
Let us put f(x) = 0
⇒ 6×2 – 7x – 3 = 0
⇒ 6×2 – 9x + 2x – 3 = 0
⇒ 3x(2x – 3) + 1(2x – 3) = 0
⇒ (2x – 3)(3x + 1) = 0
⇒ 2x – 3 = 0
x = 3/2
⇒ 3x + 1 = 0
⇒ x = -1/3
It gives us 2 zeros, for x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, for verification
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6 7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6 -1/2 = -1/2
Therefore, the relationship between zeros and their coefficients is verified.