Full solution

Q. Find two number that multiply to

140

and add to

6

Denote Numbers: Let's denote the two numbers as $x$ and $y$ . We are given two equations based on the problem statement: $\newline$ $1$ . The product of the two numbers is $140$ : $x \times y = 140$ $\newline$ $2$ . The sum of the two numbers is $6$ : $x + y = 6$ $\newline$ We need to find the values of $x$ and $y$ that satisfy both equations.
Express $y$ in terms of $x$ : From the sum equation $x + y = 6$ , we can express $y$ in terms of $x$ : $y = 6 - x$ .
Substitute into product equation: Substitute $y = 6 - x$ into the product equation $x \cdot y = 140$ to get a quadratic equation in terms of $x$ : $x \cdot (6 - x) = 140$
Expand and rewrite quadratic form: Expand the left side of the equation and write it in standard quadratic form: $\newline$ $x \times 6 - x^2 = 140$ $\newline$ $-x^2 + 6x = 140$
Set quadratic equation to zero: Move all terms to one side to set the quadratic equation to zero: $\newline$ $-x^2 + 6x - 140 = 0$
Use quadratic formula: Since the quadratic equation is not easily factorable, we can use the quadratic formula to find the values of $x$ . The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = -1$ , $b = 6$ , and $c = -140$ .
Calculate discriminant: Calculate the discriminant $\Delta = b^2 - 4ac$ : $\Delta = 6^2 - 4\cdot(-1)\cdot(-140)$ $\Delta = 36 - 560$ $\Delta = -524$ Since the discriminant is negative, there are no real solutions to the equation. This means there is a math error in our previous steps or the problem statement might be incorrect as it is not possible for two real numbers to multiply to $140$ and add to $6$ .