The next six examples will present "word problem on quadratic equation"

**Example 1**

Daniel spent $€550$ on $y$ articles, each costing $€(2x-28)$. Find x.

The number of articles is $x$

Total money spent on $x$ articles is $€550$

The cost of each article is $(2x-28)$

Hence

$x(2x-28)=550$

$2x^2-28x=550$

$2x^2-28x-550=0$

For simplicity, let's divide through by 2

$\frac{2x^2}{2}-\frac{28x}{2}-\frac{550}{2}=\frac{0}{2}$

$x^2-14x-275=0$

By factorization

$x^2-25x+11x-275=0$

$x(x-25)+11(x-25)=0$

$(x+11)(x-25)=0$

$x=-11$ or $25$

Remember that the quantity bought can not be negative, hence

$x≠-11$, $x=25$

$x=25$

If you are not familiar with quadratic equations by factorization, see my post on that here.

**Example 2**

The product of two consecutive odd numbers is 323. Find the numbers.

Let the first number be $x$

If the first number is $x$, the next odd number would be $x+2$

From the first sentence,

$x(x+2)=323$

$x^2+2x=323$

$x^2+2x-323=0$

By factorization

$x^2+19x-17x-323=0$

$x(x+19)-17(x+19)=0$

$(x-17)(x+19)=0$

$x=17$ or $-19$

An odd number is usually positive, hence, -19 is extraneous

$x=17$

If x is 17, then the next odd number is 17+2, this equal 19

Therefore, the two consecutive odd numbers are 17 and 19.

**Example 3**

If 13 times a number is subtracted from the square of the number, the result is 30. Find the number

Let the number be $x$

13 times the number means $13x$

Square of the number means $x^2$

Expressing the sentence

$x^2-13x=30$

$x^2-13x-30=0$

By factorization

$x^2-15x+2x-30=0$

$x(x-15)+2(x-15)=0$

$(x+2)(x-15)=0$

$x=-2$ or $15$

Unlike the previous examples, both answers are valid. So,

$x=-2$ or $15$

In case you want to check:

$x^2-13x=30$

When $x=-2$

$(-2)^2-13(-2)=30$

$4+26$ will give us $30$, this is correct

When $x=15$

$(15)^2-13(15)=30$

$225-195$ will give us 30, this is also correct

**Example 4**

The length of the hypotenuse side of a right-angled triangle is $2x$, the length of its opposite and adjacent sides are $(x+11)$ and $(x-3)$ respectively. Find the value of $x$.

Recalled Pythagorean's theorem

$|HYP|^2=|OPP|^2+|ADJ|^2$

Substituting the values.

$(2x)^2=(x+11)^2+(x-3)^2$

This translates to:

$4x^2=x(x+11)+11(x+11)+x(x-3)-3(x-3)$

$4x^2=x^2+11x+11x+121+x^2-3x-3x+9$

$4x^2=x^2+22x+121+x^2-6x+9$

$4x^2-x^2-22x-121-x^2+6x-9=0$

This translate into

$2x^2-16x-130=0$

For simplicity, let's Divide through by 2

$\frac{2x^2}{2}-\frac{16}{2}-\frac{130}{2}=\frac{0}{2}$

$x^2-8x-65=0$

By factorization

$x^2-13x+5x-65=0$

$x(x-13)+5(x-13)=0$

$(x+5)(x-13)=0$

$x=-5$ or $x=13$

The length of an angel can't be negative.

Hence, $-5$ is irrelevant.

$x=13$

**Example 5**

The sum of the digits of a two-digit number is $14$. The tens digit is the square of a number which is 2 less than the units digit. Find the two digits number

Let units digit by $x$

Let tens digit be $y$

From the first sentence

$x+y=14$......eqn 1

From the second sentence

$y=(x-2)^2$.....eqn 2

Substituting eqn 2 in eqn 1

$x+(x-2)^2=14$

$x+x(x-2)-2(x-2)=14$

$x+x^2-2x-2x+4=14$

$x+x^2-4x+4=14$

$x^2-4x+x+4-14=0$

$x^2-3x-10=0$

By factorization

$x^2-5x+2x-10=0$

$x(x-5)+2(x-5)=0$

$(x+2)(x-5)=0$

$x=-2$ or $x=5$

The units can not be negative, hence $x≠-2$

$x=5$

Inserting the value of x in eqn 1

$5+y=14$

$y=14-5$

$y=9$

Therefore the digits are $59$.

__Related posts__**Example 6**

Two numbers differ by $7$. If their product is $120$, find the numbers

Let the first number be $x$

Let the second number be $y$

$x-y=7$..........eqn 1

$xy=120$......eqn 2

Re-arranging eqn 1

$x=7+y$.....eqn 3

Substituting eqn 3 in eqn 2

$(7+y)y=120$

$7y+y^2=120$

$y^2+7y-120=0$

By factorization

$y^2+15y-8y-120=0$

$y(y+15)-8(y+15)=0$

$(y-8)(y+15)=0$

$y=8$ or $-15$

Inserting the value of y in Eqn 3

When $y=8$

$x=7+8$

$x=15$

When $y=-15$

$x=7+(-15)$

$x=7-15$

$x=-8$

There you have it, you should be able to solve any question on the quadratic equation. But still, if you have got questions regarding the quadratic equation, be sure to ask our Facebook community.

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