Drag the factors to the correct locations on the image. Not all factors will be used.What is the factored form of this expressio
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1 answer:
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Answer:
1) Real and same.
2) Real and distinct
3) Real and distinct
4) Real and distinct
5) Real and distinct
6) Real and distinct
7) Real and distinct
8) Real and distinct
Step-by-step explanation:
If a quadratic equation is in the form of ax² + bx + c = 0, then Discriminant of the equation D = b² - 4ac, which governs the nature of roots the equation has.
If D > 0, then there will be two different real roots.
If D = 0, then two same and real roots.
If D < 0, then two distinct but imaginary roots.
Now, 1) x² + 6x + 9 = 0 has D = 6² - 4 × 9 × 1 = 0. So, there will be two same and real roots.
2) 5x² - x = 4x² + 2
⇒ x² - x - 2 = 0
It has D = (-1)² - 4 × 1 × (-2) = 9. Therefore, the roots will be two distinct and real.
3)
⇒ 5 = x² - 4x
⇒ x² - 4x - 5 = 0.
So, D = (-4)² - 4 × (1) × (- 5) = 36. So, the equation has two real and distinct rools.
4) x(x - 5) = 4(5 - x)
⇒ x² - x - 20 = 0
Hence, D = (-1)² - 4 × 1 × (-20) = 81
So, the roots will be real and distinct.
5) x² + 7x + 1 = 0 has D = 7² - 4 × 1 × 1 = 45
So, the roots will be real and distinct.
6) 2x² + 9x + 3 = 0, has D = 9² - 4 × 2 × 3 = 57.
Hence, the roots will be real and distinct.
7) 5x² - 6 = 13x
⇒ 5x² - 13x - 6 = 0
So, D = (-13)² - 4 × 5 × (-6) = 289
So, the roots will be real and distinct.
8) x² - x = 3(x + 7)
⇒ x² - 4x - 21 = 0
It has D = (-4)² - 4 × 1 × (-21) = 100
So, the roots will be real and distinct. (Answer)
Answer:
Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.
<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that . Thus, A↔A.
<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that . In this equality we can perform a right multiplication by
and obtain
. Then, in the obtained equality we perform a left multiplication by P and get
. If we write
and
we have
. Thus, B↔A.
<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have and from B↔C we have
. Now, if we substitute the last equality into the first one we get
.
Recall that if P and Q are invertible, then QP is invertible and . So, if we denote R=QP we obtained that
. Hence, A↔C.
Therefore, the relation is an <em>equivalence relation</em>.