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2 years ago

Solve x2 – 12x + 59 = 0 for x.

1 answer:

3 0

59 is a tough bird to deal with; its only factors are 1 and 59.

Thus, forget about factoring. Instead, use the quadratic formula, or solve the equation by completing the square.

Please note: x2 is ambiguous. Please write x^2 to indicate "the square of 2."

Here you have 1x^2 - 12x + 59 = 0, for which a=1, b=-12 and c=59.

Use the quadratic formula: x=[-b plus or minus sqrt(b^2-4ac)] / (2a)

to find the two roots. Notice that the "discriminant" b^2 - 4ac will be negative, meaning that your two roots will be "complex."

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2 years ago

What is the product of -3 and 5?

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2 years ago

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3 years ago

Read 2 more answers

Question 1

drek231 [11]

QUESTION 1

We want to expand $(x-2)^6$ .

We apply the binomial theorem which is given by the formula

$(a+b)^n=^nC_0a^nb^0+^nC_1a^{n-1}b^1+^nC_2a^{n-2}b^2+...+^nC_na^{n-n}b^n$ .

By comparison,

$a=x,b=-2,n=6$ .

We substitute all these values to obtain,

$(x-2)^6=^6C_0x^6(-2)^0+^6C_1x^{6-1}(-2)^1+^6C_2x^{6-2}(-2)^2+^6C_3x^{6-3}(-2)^3+^6C_4x^{6-4}(-2)^4+^6C_5x^{6-5}(-2)^5+^6C_6x^{6-6}(-2)^6$ .

We now simplify to obtain,

$(x-2)^6=^nC_0x^6(-2)^0+^6C_1x^{5}(-2)^1+^6C_2x^{4}(-2)^2+^6C_3x^{3}(-2)^3+^6C_4x^{2}(-2)^4+^6C_5x^{1}(-2)^5+^6C_6x^{0}(-2)^6$ .

This gives,

$(x-2)^6=x^6-12x^{5}+60x^{4}-160x^{3}(-2)^3+240x^{2}-1925x+64$ .

Ans:C

QUESTION 2

We want to expand

$(x+2y)^4$ .

We apply the binomial theorem to obtain,

$(x+2y)^4=^4C_0x^4(2y)^0+^4C_1x^{4-1}(2y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(2y)^3+^4C_4x^{4-4}(2y)^4$ .

We simplify to get,

$(x+2y)^4=x^4(2y)^0+4x^{3}(2y)^1+6x^{2}(2y)^2+4x^{1}(2y)^3+x^{0}(2y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4+8x^{3}y+24x^{2}y^2+32x^{1}y^3+16y^4$

Ans:B

QUESTION 3

We want to find the number of terms in the binomial expansion,

$(a+b)^{20}$ .

In the above expression, $n=20$ .

The number of terms in a binomial expression is $(n+1)=20+1=21$ .

Therefore there are 21 terms in the binomial expansion.

Ans:C

QUESTION 4

We want to expand

$(x-y)^4$ .

We apply the binomial theorem to obtain,

$(x-y)^4=^4C_0x^4(-y)^0+^4C_1x^{4-1}(-y)^1+^4C_2x^{4-2}(2y)^2+^4C_3x^{4-3}(-y)^3+^4C_4x^{4-4}(-y)^4$ .

We simplify to get,

$(x+2y)^4=^x^4(-y)^0+4x^{3}(-y)^1+6x^{2}(-y)^2+4x^{1}(-y)^3+x^{0}(-y)^4$ .

We simplify further to obtain,

$(x+2y)^4=x^4-4x^{3}y+6x^{2}y^2-4x^{1}y^3+y^4$

Ans: C

QUESTION 5

We want to expand $(5a+b)^5$

We apply the binomial theorem to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{5-1}(b)^1+^5C_2(5a)^{5-2}(b)^2+^5C_3(5a)^{5-3}(b)^3+^5C_4(5a)^{5-4}(b)^4+^5C_5(5a)^{5-5}(b)^5$ .

We simplify to obtain,

$(5a+b)^5=^5C_0(5a)^5(b)^0+^5C_1(5a)^{4}(b)^1+^5C_2(5a)^{3}(b)^2+^5C_3(5a)^{2}(b)^3+^5C_4(5a)^{1}(b)^4+^5C_5(5a)^{0}(b)^5$ .

This finally gives us,

$(5a+b)^5=3125a^5+3125a^{4}b+1250a^{3}b^2+^250a^{2}(b)^3+25a(b)^4+b^5$ .

Ans:B

QUESTION 6

We want to expand $(x+2y)^5$ .

We apply the binomial theorem to obtain,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{5-1}(2y)^1+^5C_2(x)^{5-2}(2y)^2+^5C_3(x)^{5-3}(2y)^3+^5C_4(x)^{5-4}(2y)^4+^5C_5(x)^{5-5}(2y)^5$ .

We simplify to get,

$(x+2y)^5=^5C_0(x)^5(2y)^0+^5C_1(x)^{4}(2y)^1+^5C_2(x)^{3}(2y)^2+^5C_3(x)^{2}(2y)^3+^5C_4(x)^{1}(2y)^4+^5C_5(x)^{0}(2y)^5$ .

This will give us,

$(x+2y)^5=x^5+^10(x)^{4}y+40(x)^{3}y^2+80(x)^{2}y^3+80(x)y^4+32y^5$ .

Ans:A

QUESTION 7

We want to find the 6th term of $(a-y)^7$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=7,b=-y$

We substitute to obtain,

$T_{5+1}=^7C_5a^{7-5}(-y)^5$ .

$T_{6}=-21a^{2}y^5$ .

Ans:D

QUESTION 8.

We want to find the 6th term of $(2x-3y)^{11}$

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=11,a=2x,b=-3y$

We substitute to obtain,

$T_{5+1}=^{11}C_5(2x)^{11-5}(-3y)^5$ .

$T_{6}=-7,185,024x^{6}y^5$ .

Ans:D

QUESTION 9

We want to find the 6th term of $(x+y)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=5,n=8,a=x,b=y$

We substitute to obtain,

$T_{5+1}=^8C_5(x)^{8-5}(y)^5$ .

$T_{6}=56a^{3}y^5$ .

Ans: A

We want to find the 7th term of $(x+4)^8$ .

The nth term is given by the formula,

$T_{r+1}=^nC_ra^{n-r}b^r$ .

Where $r=6,n=8,a=x,b=4$

We substitute to obtain,

$T_{6+1}=^8C_5(x)^{8-6}(4)^6$ .

$T_{7}=114688x^{2}$ .

Ans:A

4 0

2 years ago

3/5 as a rational number

zhenek [66]

<span>3/5 is already a rational number :)</span>

6 0

3 years ago

Read 2 more answers