The expression shown below is a difference of two squares.
<h3>Is a given expression a perfect square trinomial or a difference of two squares?</h3>
In this problem we have an algebraic expression that has to be checked by algebraic procedures. The complete procedure is shown below:
(x² + 8 · x + 16) · (x² - 8 · x + 16) Given
(x + 4)² · (x - 4)² Perfect square trinomial
[(x + 4) · (x - 4)] · [(x + 4) · (x - 4)] Definition of power / Associative and commutative property
(x² - 16)² Difference of squares / Definition of power / Result
The expression shown below is a difference of two squares.
To learn more on differences of squares: brainly.com/question/11801811
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Answer:
Roots are not real
Step-by-step explanation:
To prove : The roots of x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0 are real for all real values of k ?
Solution :
The roots are real when discriminant is greater than equal to zero.
i.e. b^2-4ac\geq 0b
2
−4ac≥0
The quadratic equation x^2 +(1-k)x+k-3=0x
2
+(1−k)x+k−3=0
Here, a=1, b=1-k and c=k-3
Substitute the values,
We find the discriminant,
D=(1-k)^2-4(1)(k-3)D=(1−k)
2
−4(1)(k−3)
D=1+k^2-2k-4k+12D=1+k
2
−2k−4k+12
D=k^2-6k+13D=k
2
−6k+13
D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))
For roots to be real, D ≥ 0
But the roots are imaginary therefore the roots of the given equation are not real for any value of k.
Answer:
b,f
Step-by-step explanation:
b: if sandi scored 3 more points than randi, then sandis points are equal to 3+randi
f: so the equation is 3+r=s,
so input values
3+r=18
r=18-3
r=15