How to simplify 6 +8b^2 +3b -6 - 10b^2
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<h2>Perfect Squares</h2>
Perfect square formula/rules:
Trinomials are often organized like .
The <em>b</em> value in this case is <em>c</em>, and it will always equal the square of half of the <em>b</em> value.
- Perfect square trinomial:
- or
We're given:
In a trinomial, we're given the and
values. <em>a</em> in this case is 1 and <em>b</em> in this case is 4. To find the third value by dividing 4 by 2 and squaring the quotient:
- 4 ÷ 2 = 2
- 2² = 4
Therefore, the term that we can add is + 4.
To write this as the square of a bracketed expression, we can follow the rule :
(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
See the figure for the graph:
(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )
∴ x → -x hence '-1' is the eigen value.
∴ y → y hence '1' is the eigen value.
also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.
( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.
(b) ∵ T(x, y) = (-x, y)
T(x) = -x = (-1)(x) + 0(y)
T(y) = y = 0(x) + 1(y)
Matrix Representation of
now, eigen value of 'T'
after solving the determinant,
we get two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
Hence,
(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.
(b) two eigen values of 'k' = 1, -1
for k = 1, eigen vector is
for k = -1 eigen vector is
Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052
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