Answer :
Let's solve each question step-by-step.
### Question 1
The product of a number and the number increased by 3 is 54. Find the two possible values the number could be.
Let's denote the number by [tex]\( x \)[/tex].
The product of [tex]\( x \)[/tex] and [tex]\( x + 3 \)[/tex] is given by the equation:
[tex]\[ x(x + 3) = 54 \][/tex]
Expanding this equation, we get:
[tex]\[ x^2 + 3x = 54 \][/tex]
Rearranging it into a standard quadratic equation form:
[tex]\[ x^2 + 3x - 54 = 0 \][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -54 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 216}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{225}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 15}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{-3 + 15}{2} = \frac{12}{2} = 6 \)[/tex]
2. [tex]\( x = \frac{-3 - 15}{2} = \frac{-18}{2} = -9 \)[/tex]
So, the two possible values for the number are [tex]\( 6 \)[/tex] and [tex]\( -9 \)[/tex].
### Question 2
The difference between a number and its square is 42. Find the number.
Let's denote the number by [tex]\( x \)[/tex].
The equation is given by:
[tex]\[ x - x^2 = 42 \][/tex]
Rearranging the equation into a standard quadratic form:
[tex]\[ -x^2 + x - 42 = 0 \][/tex]
or
[tex]\[ x^2 - x + 42 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -42 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-42)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 168}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 13}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{1 + 13}{2} = \frac{14}{2} = 7 \)[/tex]
2. [tex]\( x = \frac{1 - 13}{2} = \frac{-12}{2} = -6 \)[/tex]
So, the two possible values for the number are [tex]\( 7 \)[/tex] and [tex]\( -6 \)[/tex].
### Question 3
When 15 is added to the square of a number, the result is eight times the original number. Find the number.
Let's denote the number by [tex]\( x \)[/tex].
The equation is given by:
[tex]\[ x^2 + 15 = 8x \][/tex]
Rearranging the equation into a standard quadratic form:
[tex]\[ x^2 - 8x + 15 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 15 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 - 60}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm 2}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{8 + 2}{2} = \frac{10}{2} = 5 \)[/tex]
2. [tex]\( x = \frac{8 - 2}{2} = \frac{6}{2} = 3 \)[/tex]
So, the two possible values for the number are [tex]\( 5 \)[/tex] and [tex]\( 3 \)[/tex].
### Question 1
The product of a number and the number increased by 3 is 54. Find the two possible values the number could be.
Let's denote the number by [tex]\( x \)[/tex].
The product of [tex]\( x \)[/tex] and [tex]\( x + 3 \)[/tex] is given by the equation:
[tex]\[ x(x + 3) = 54 \][/tex]
Expanding this equation, we get:
[tex]\[ x^2 + 3x = 54 \][/tex]
Rearranging it into a standard quadratic equation form:
[tex]\[ x^2 + 3x - 54 = 0 \][/tex]
To solve this quadratic equation, we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -54 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 + 216}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{225}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm 15}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{-3 + 15}{2} = \frac{12}{2} = 6 \)[/tex]
2. [tex]\( x = \frac{-3 - 15}{2} = \frac{-18}{2} = -9 \)[/tex]
So, the two possible values for the number are [tex]\( 6 \)[/tex] and [tex]\( -9 \)[/tex].
### Question 2
The difference between a number and its square is 42. Find the number.
Let's denote the number by [tex]\( x \)[/tex].
The equation is given by:
[tex]\[ x - x^2 = 42 \][/tex]
Rearranging the equation into a standard quadratic form:
[tex]\[ -x^2 + x - 42 = 0 \][/tex]
or
[tex]\[ x^2 - x + 42 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -42 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-42)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 168}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{169}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 13}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{1 + 13}{2} = \frac{14}{2} = 7 \)[/tex]
2. [tex]\( x = \frac{1 - 13}{2} = \frac{-12}{2} = -6 \)[/tex]
So, the two possible values for the number are [tex]\( 7 \)[/tex] and [tex]\( -6 \)[/tex].
### Question 3
When 15 is added to the square of a number, the result is eight times the original number. Find the number.
Let's denote the number by [tex]\( x \)[/tex].
The equation is given by:
[tex]\[ x^2 + 15 = 8x \][/tex]
Rearranging the equation into a standard quadratic form:
[tex]\[ x^2 - 8x + 15 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = -8 \)[/tex], and [tex]\( c = 15 \)[/tex].
Plugging in these values:
[tex]\[ x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 15}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{64 - 60}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm \sqrt{4}}{2} \][/tex]
[tex]\[ x = \frac{8 \pm 2}{2} \][/tex]
This yields two solutions:
1. [tex]\( x = \frac{8 + 2}{2} = \frac{10}{2} = 5 \)[/tex]
2. [tex]\( x = \frac{8 - 2}{2} = \frac{6}{2} = 3 \)[/tex]
So, the two possible values for the number are [tex]\( 5 \)[/tex] and [tex]\( 3 \)[/tex].
