Find the number of roots for each equation.
5. 5x4 + 12x3 – x2 + 3x + 5 = 0

The number of roots for  equation is 4 .

Step-by-step explanation:

Here, the given function polynomial is :

The Fundamental Theorem of Algebra says that a polynomial of degree n will have exactly n roots (counting multiplicity).

Now here, the degree if the polynomial is 4 (highest power of variable x).

So, according to the Fundamental Theorem, the given polynomial can have AT MOST 4 roots, counting Multiplicity.

Hence,  the number of roots for  equation is 4 .

3. = 100 A.10

Step-by-step explanation:

10 x 10 =100

Answer and Step-by-step explanation:

A root is considered a real root if it is not an imaginary number (a number in the form of a + bi, where i is a number that is not on the real number line, hence the name: imaginary). Let's inspect the given polynomials.

1. (x-4)(x+3)^2(x-1)^3=0, the equation implies

x – 4 = 0 or (x + 3)^2 = 0 or (x – 1)^3 = 0, this gives:

x = 4

x = –3 twice

x = 1 thrice

They are all real numbers, therefore it has 6 real roots.

2. x^2(x^3-1)=0

x² = 0

x³ – 1 = x³ – 1³ = (x – 1)(x² + x + 1) = 0

The quadratic expressions gives 2 imaginary roots. To check this we can use the determinant formula.

D = b² – 4ac = 1² – 4(1)(1) = 1 – 4 = –3 < 0. This gives roots that are not real.

Therefore the real roots of this polynomial are: 0(twice) and 1. Hence, it has 3 real roots.

3. x(x+3)(x-6)^2=0

x = 0

x = 3

x = 6 twice

Thus, it has 4 real roots.

4. 3x(x^3-1)^2=0

3x = 0 and this gives x = 0

(x^3 – 1)² = [(x – 1)(x² + x + 1)]² = (x – 1)²(x² + x + 1)² = 0, only two roots are real here (the quadratic expression has unreal roots, therefore 1 twice are the real roots).

Therefore, this has 3 real roots.

5. (x^3-8)(x^4+1)=0

x³ – 8 = x³ – 2³ = (x – 2)(x² + 2x + 4) = 0

The only real root here is 2, the quadratic expression had imaginary roots and can be checked using the determinant formula

D = b² – 4ac = 2² – 4(1)(4) = 4 – 16 = – 12 < 0.

x⁴ + 1 = 0 has imaginary roots. Therefore, this polynomial has only one real root.

Answer and Step-by-step explanation:

A root is considered a real root if it is not an imaginary number (a number in the form of a + bi, where i is a number that is not on the real number line, hence the name: imaginary). Let's inspect the given polynomials.

1. (x-4)(x+3)^2(x-1)^3=0, the equation implies

x – 4 = 0 or (x + 3)^2 = 0 or (x – 1)^3 = 0, this gives:

x = 4

x = –3 twice

x = 1 thrice

They are all real numbers, therefore it has 6 real roots.

2. x^2(x^3-1)=0

x² = 0

x³ – 1 = x³ – 1³ = (x – 1)(x² + x + 1) = 0

The quadratic expressions gives 2 imaginary roots. To check this we can use the determinant formula.

D = b² – 4ac = 1² – 4(1)(1) = 1 – 4 = –3 < 0. This gives roots that are not real.

Therefore the real roots of this polynomial are: 0(twice) and 1. Hence, it has 3 real roots.

3. x(x+3)(x-6)^2=0

x = 0

x = 3

x = 6 twice

Thus, it has 4 real roots.

4. 3x(x^3-1)^2=0

3x = 0 and this gives x = 0

(x^3 – 1)² = [(x – 1)(x² + x + 1)]² = (x – 1)²(x² + x + 1)² = 0, only two roots are real here (the quadratic expression has unreal roots, therefore 1 twice are the real roots).

Therefore, this has 3 real roots.

5. (x^3-8)(x^4+1)=0

x³ – 8 = x³ – 2³ = (x – 2)(x² + 2x + 4) = 0

The only real root here is 2, the quadratic expression had imaginary roots and can be checked using the determinant formula

D = b² – 4ac = 2² – 4(1)(4) = 4 – 16 = – 12 < 0.

x⁴ + 1 = 0 has imaginary roots. Therefore, this polynomial has only one real root.

3. = 100 A.10

Step-by-step explanation:

10 x 10 =100

Given the quadratic equation ax^2 + bx + c = 0, the coefficients of x^2, x and 1 are a, b and c respectively.  

(a) and (b) are not valid quadratics, because they omit the exponent 2.   Rewrite these as comments within your original post, please.

(c) 4x^2+20x+25=0:  Here, a = 4, b = 20 and c = 25.

24.6967 meters

Step-by-step explanation:

The roots of the tree go 6 and 5 over 12 meters below the ground level.

Now, 6 and 5 over 12 meters is equivalent to 6.4167 meters.

Again the top of the tree is 18.28 meters high from the ground level.

Therefore, the total height of the tree from the bottom of the root to the top is  

(6.4167 +18.28) = 24.6967 meters (Answer)

determinant:

(a)

D<0 means there are no real roots. there are two complex roots with imaginary components.

(b) D=16+20=36>0

D>0 means there are two real roots

(c) D = 20^2-4*4*25 = 0

D=0 means there is one real root with multiplicity 2