Home Mathematics Need on these two cause i’m very lost on them Need on these two cause i’m very lost on themMathematics Spooderfaxy7813October 23, 20213 CommentsNeed on these two cause i’m very lost on them[tex]Need on these two cause i’m very lost on them[/tex]
see explanationStep-by-step explanation:Givenf(x) = (x - 4)(2x - 1)²(x - 2)²To find the roots equate f(x) to zero, that is(x - 4)(2x- 1)²(x - 2)² = 0Equate each of the factors to zero and solve for xx - 4 = 0 ⇒ x = 42x - 1 = 0 ⇒ x = [tex]\frac{1}{2}[/tex] ← with multiplicity 2x - 2 = 0 ⇒ x = 2 ← with multiplicity 2Hence the roots are{ 4, [tex]\frac{1}{2}[/tex], 2 }Givenf(x) = x³ + 4x² + 7x + 6Note thatf(- 2) = (- 2)³ + 4(- 2)² + 7(- 2) + 6 = - 8 + 16 - 14 + 6 = 0Since f(- 2) = 0 then by the factor theorem x = - 2 is a root and (x + 2) a factorUsing synthetic division- 2 | 1 4 7 6 - 2 - 4 - 6-------------- 1 2 3 0Thusf(x) = (x + 2)(x² + 2x + 3)Solve x² + 2x + 3 using the quadratic formulawith a = 1, b = 2 and c = 3x = (- 2 ± [tex]\sqrt{2^2-(4(1)(3)}[/tex] ) / 2 = ( - 2 ± [tex]\sqrt{4-12}[/tex] ) / 2 = ( - 2 ± [tex]\sqrt{-8}[/tex] ) / 2 = (- 2 ± 2i[tex]\sqrt{2}[/tex] ) / 2 = - 1 ± i[tex]\sqrt{2}[/tex]Hence roots are{ - 2, - 1 + i[tex]\sqrt{2}[/tex], - 1 - i[tex]\sqrt{2}[/tex] } Reply
Dis the correct answer
215.92 dollars is the answer
see explanation
Step-by-step explanation:
Given
f(x) = (x - 4)(2x - 1)²(x - 2)²
To find the roots equate f(x) to zero, that is
(x - 4)(2x- 1)²(x - 2)² = 0
Equate each of the factors to zero and solve for x
x - 4 = 0 ⇒ x = 4
2x - 1 = 0 ⇒ x = [tex]\frac{1}{2}[/tex] ← with multiplicity 2
x - 2 = 0 ⇒ x = 2 ← with multiplicity 2
Hence the roots are
{ 4, [tex]\frac{1}{2}[/tex], 2 }
Given
f(x) = x³ + 4x² + 7x + 6
Note that
f(- 2) = (- 2)³ + 4(- 2)² + 7(- 2) + 6 = - 8 + 16 - 14 + 6 = 0
Since f(- 2) = 0 then by the factor theorem x = - 2 is a root and (x + 2) a factor
Using synthetic division
- 2 | 1 4 7 6
- 2 - 4 - 6
--------------
1 2 3 0
Thus
f(x) = (x + 2)(x² + 2x + 3)
Solve x² + 2x + 3 using the quadratic formula
with a = 1, b = 2 and c = 3
x = (- 2 ± [tex]\sqrt{2^2-(4(1)(3)}[/tex] ) / 2
= ( - 2 ± [tex]\sqrt{4-12}[/tex] ) / 2
= ( - 2 ± [tex]\sqrt{-8}[/tex] ) / 2
= (- 2 ± 2i[tex]\sqrt{2}[/tex] ) / 2
= - 1 ± i[tex]\sqrt{2}[/tex]
Hence roots are
{ - 2, - 1 + i[tex]\sqrt{2}[/tex], - 1 - i[tex]\sqrt{2}[/tex] }