
To find the number of zeros.

Answer to Problem 7CA
No zerosOne zeroTwo zerosf(x)=3x2+4x+2f(x)=−6x2−5f(x)=4x2−8x+4f(x)=7x2f(x)=−x2+2xf(x)=x2−3x−21
Explanation of Solution
Given information: f(x)=3x2+4x+2 , f(x)=4x2−8x+4 , f(x)=7x2 , f(x)=−x2+2x , f(x)=x2−3x−21 and f(x)=−6x2−5
Formula Used: D=b2−4ac
Calculation:
Discriminant be used to confirm the number of real solutions of a quadratic equation because discriminant method is much easier and shorter than completing the square method.
We can easily calculate discriminant by D = b2−4ac and can know how many real numbers equation has.
- If D equal to negative then x has no real solution.
- If D equal to 0 then x has 1 real solution.
- If D equal to positive then x has 2 real solution.
(1)
f(x)=3x2+4x+2
As, D = b2−4ac
Now, a=3 , b=4 and c=2
D = 42−4×3×2
So,
D = 16−24
Hence,
D =−8
D is equal to negative then x has no real solution.
(2)
f(x)=4x2−8x+4
As, D = b2−4ac
Now, a=4 , b=−8 and c=4
D = (−8)2−4×4×4
So,
D = 64−64
Hence,
D =0
D is equal to 0 then x has 1 real solution.
(3)
f(x)=7x2
As, D = b2−4ac
Now, a=7 , b=0 and c=0
D = (0)2−4×7×0
So,
D = 0−0
Hence,
D =0
D is equal to 0 then x has 1 real solution.
(4)
f(x)=−x2+2x
As, D = b2−4ac
Now, a=−1 , b=2 and c=0
D = (2)2−4×(−1)×0
So,
D = 4−0
Hence,
D =4
D is equal to positive then x has 2 real solution.
(5)
f(x)=x2−3x−21
As, D = b2−4ac
Now, a=1 , b=−3 and c=−21
D = (−3)2−4×1×(−21)
So,
D = 9+84
Hence,
D =93
D is equal to positive then x has 2 real solution.
(6)
f(x)=−6x2−5
As, D = b2−4ac
Now, a=−6 , b=0 and c=−5
D = (0)2−4×(−6)×(−5)
So,
D = 0−120
Hence,
D =−120
D is equal to negative then x has no real solution.
Hence Solution,
No zerosOne zeroTwo zerosf(x)=3x2+4x+2f(x)=−6x2−5f(x)=4x2−8x+4f(x)=7x2f(x)=−x2+2xf(x)=x2−3x−21
Chapter 10 Solutions
BIG IDEAS MATH Integrated Math 1: Student Edition 2016
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