Question

Model the given statement in algebraic equation, the given statement is “the product of “8” and the sum of a number “x” and “3”?

Answer 1

For converting a statement into an algebraic equation you have to assume variables According to the statement given, the variable assumed is purely dependent on the quantities asked for, and after that just make the mathematical expression according to the statement and after solving the algebraic equation you can find the value of the variable you assumed.

Formulae Used:
$D(discriminant) = \sqrt {{b^2} - 4ac}$,

$$roots\,of\,equation\,are: \\\ = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\$$

Complete step by step solution:
For the given statement we can get that the question is asking for a variable say “y” such that

$$8 = x \times y \\\ y = x + 3 \\\$$

Now rearranging the above first equation $8 = x \times y$ we get;
$\dfrac{8}{x} = y$

Now putting value of “y” in equation $y = x + 3$ we get;

$$\dfrac{8}{x} = x + 3 \\\ x - \dfrac{8}{x} = - 3 \\\ \dfrac{{{x^2} - 8}}{x} = 3 \\\ {x^2} - 8 = 3x \\\ {x^2} - 3x = 8 \\\$$

Now solving above quadratic equation we get;

$$comparing\,with\varepsilon \,general\,quadratic\,equation\,a{x^2} + bx + c\,\,we\,get \\\ a = 1,\,b = - 3,\,c = - 8 \\\ u\sin g\,Shreedharacharya\,rule\, \\\ D(discriminant) = \sqrt {{b^2} - 4ac} = \sqrt {{{( - 3)}^2} - 4 \times 1 \times ( - 8)} \\\ = \sqrt {9 + 32} \\\ = \sqrt {41} \\\$$ $$now,\,roots\,of\,equation\,are: \\\ = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\ = \dfrac{{ - ( - 3) + \sqrt {41} }}{{2(1)}},\,\dfrac{{ - ( - 3) - \sqrt {41} }}{{2(1)}}\,(D = \sqrt {{b^2} - 4ac} ) \\\ = \dfrac{{3 + \sqrt {41} }}{2},\,\dfrac{{3 - \sqrt {41} }}{2} \\\ = \dfrac{{3 + 6.40}}{2},\,\dfrac{{3 - 6.40}}{2}\,(\sqrt {41} = 6.40) \\\ = \dfrac{{9.40}}{2},\,\dfrac{{ - 3.40}}{2} \\\ = 4.7, - 1.7 \\\$$

Now we see that in the given question the product is non- negative hence the negative hence the
a negative term obtained as a root of the quadratic equation is not taken and only the positive term is accepted.

So or required value for “x” is $4.7$

Now value of “y” equals to:

$$8 = x \times y \\\ y = \dfrac{8}{x} \\\ y = \dfrac{8}{{4.7}},\,(x = 4.7) \\\ y = 1.70 \\\$$

Hence the required number is $1.70$

Note: The conversion of statement to the algebraic equation needs only the understanding of the statement, you should be careful about the words written over there and accordingly you can re-write it in the terms of math’s. For example “of” is used for product, “ratio” is used for division, “sum” is used for addition and “difference” is used for subtraction.