10 Quadratic Equations Word Problems

We can use the letters X (smaller number) and Y (larger number) to represent both numbers. Then, from the statement, we can obtain the following equations:

$latex X+Y=17~~[1]$

$latex XY=60~~[2]$

Now, we can rearrange equation 1 to get $latex Y=17-X$. Substituting this expression into the second equation, we have:

$latex X(17-X)=60$

Expanding the parentheses and rearranging the equation, we have:

$latex 17X-X^2=60$

$latex X^2-17X+60=0$

Now, we can factor the equation to solve:

$latex X^2-17X+60=0$

$latex (X-12)(X-5)=0$

$latex X-12=0~~$ or $latex ~~X-5=0$

$latex X=12~~$ or $latex ~~X=5$

Therefore, we have:

• If $latex X=5$, we have $latex Y=17-5=12$. This solution is correct because X<Y.
• If $latex X=12$, we have $latex Y=17-12=5$

The numbers are 12 and 5.

Using the information in the statement and using the letters X (smaller number) and Y (larger number) to represent the numbers, we can obtain the following equations:

$latex Y-X=5 ~~[1]$

$latex XY=126~~[2]$

Now, we can write the first equation as $latex Y=5+X$ and substituting it into the second equation, we have:

$latex X(5+X)=126$

$latex 5X+X^2=126$

$latex X^2+5X-126=0$

Now, we can factor and solve:

$latex (X+13)(X-9)=0$

$latex X+13=0~~$ or $latex ~~X-9=0$

$latex X=-13~~$ or $latex ~~X=9$

Therefore, we have:

• If $latex X=-13$, we have $latex Y=5-13=-8$
• If $latex X=9$, we have $latex Y=5+9=14$. This solution is correct because $latex XY=126$

The numbers are 12 and 5.

We can represent the shorter side with x. This means that the longest side is equal to x+7.

Also, we remember that the area of a rectangle is equal to the product of its sides. Therefore, we can form the following equation:

$latex x(x+7)=78$

Expanding and writing the equation in the form $late ax^2+bx+c=0$, we have:

$latex x^2+7x-78=0$

Now, we can use factorization to solve:

$latex (x+13)(x-6)=0$

$latex x+13=0~~$ or $latex ~~x-6=0$

$latex x=-13~~$ or $latex ~~x=6$

Since we can’t have negative lengths, we have $latex x=6$, so the lengths are 6 and 13.

Using the letters X (short side) and Y (long side) to represent the sides, we can obtain the following equations using the given information:

$latex Y=2X~~[1]$

$latex XY=200~~[2]$

Now, we can substitute equation 1 into equation 2 and we have:

$latex X(2X)=200$

$latex 2X^2=200$

To solve this equation, we have to isolate X² completely and take the square root of both sides:

$latex X^2=100$

$latex X=\pm\sqrt{100}$

$latex X=\pm 10$

Since a length can’t be negative, we have $latex X=10$. Therefore, we have $latex Y=20$.

Representing the first number with x, we can deduce that a consecutive odd number is equal to $latex x+2$. Therefore, using the information in the statement, we form an equation with the squares of the consecutive numbers:

$latex (x+2)^2-x^2=48$

Expanding, simplifying and solving, we have:

$latex x^2+4x+4-x^2=48$

$latex 4x+4=48$

$latex 4x=44$

$latex x=11$

The consecutive odd numbers are 11 and 13.

Since the triangle is a right triangle, we can use the Pythagorean theorem to find an equation for the lengths of the three sides of the triangle.

The statement tells us that 10 is the hypotenuse of the triangle. Therefore, we have:

$latex x^2+(x+2)^2=10^2$

Expanding and simplifying the equation, we have:

$latex x^2+(x+2)^2=10^2$

$latex x^2+x^2+4x+4=100$

$latex 2x^2+4x+4=100$

$latex 2x^2+4x-96=0$

Now, we can divide the entire equation by 2 to simplify it:

$latex x^2+2x-48=0$

Solving by factorization, we have:

$latex (x+8)(x-6)=0$

$latex x+8=0~~$ or $latex ~~x-6=0$

$latex x=-8~~$ or $latex ~~x=6$

Since we can’t have a negative length, the answer is $latex x=6$.

We can represent one of the numbers with x. This means that the other number is equal to 48/x since their product is equal to 48.

Therefore, considering that the average of two numbers is equal to the sum of the numbers divided by 2, we have:

$latex \frac{x+48/x}{2}=7$

Simplifying, we have:

$latex x+\frac{48}{x}=14$

$latex x^2+48=14x$

$latex x^2-14x+48=0$

Now, we can solve by factorization:

$latex (x-8)(x-6)=0$

$latex x-8=0~~$ or $latex ~~x-6=0$

$latex x=8~~$ or $latex ~~x=6$

The numbers are 6 and 8.

In this problem, there are two areas that we need to consider, the area of the original square and the area of the larger square (length +4).

Using the x to represent the sides of the original square and remembering that the area of a square is equal to the length of one of the sides squared, we have:

Larger area = 9 × Original area

$latex (x+4)^2=9x^2$

Expanding and simplifying, we have:

$latex x^2+8x+16=9x^2$

$latex 8x^2-8x-16=0$

We can divide the entire equation by 8 and solving by factoring, we have:

$latex x^2-x-2=0$

$latex (x-2)(x+1)=0$

$latex x-2=0~~$ or $latex ~~x+1=0$

$latex x=2~~$ or $latex ~~x=-1$

Since a length cannot be negative, the answer is $latex x=2$.

From the statement, we can deduce the following lengths of the triangle:

• Short side = x-2
• Intermediate side = x (2 units difference with short side)
• Hypotenuse = x+2 (4 units more than short side)

Now, we can use the Pythagorean theorem to write an equation for the three lengths:

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

Expanding and simplifying, we have:

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

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

$latex x^2-8x=0$

$latex x(x-8)=0$

$latex x=0~~$ or $latex ~~x=8$

Since the length cannot be 0, we have $latex x=8$.

We can divide the figure into two rectangles, one rectangle has the area 7x and the other rectangle has the area x(2x+3). Adding these areas, we have:

$latex 7x+x(2x+3)=100$

Expanding and simplifying, we have:

$latex 7x+2x^2+3x=100$

$latex 2x^2+10x-100=0$

Now, we can divide the entire equation by 2 and solve by factoring:

$latex x^2+5x-50=0$

$latex (x+10)(x-5)=0$

$latex x=-10~~$ or $latex ~~x=5$

Since we can’t have a negative length, the answer is $latex x=5$.