# Solving Word Problems Involving Quadratic Equations

**Solving Word Problems Involving Quadratic Equation**

**One of the most important topics in Algebra is solving quadratic equations. It has found wide applications in many areas of Mathematics. This hub presents several problems involving quadratic equations with their solution. Hope you will enjoy this.**

**Sample Problem One :**

**The sum of two numbers is 18 and the sum of their square is 170. Find the numbers.**

**Solution:**

**Let X = the first number**

**18 – X = the second number**

**X^2 = square of the first number**

**(18-X)^2 = square of the second number**

**Working Equation:**

**X^2 + (18-X)^2 = 170**

**X^2 + 324 -36X + X^2 = 170**

**2X^2 -36X +324 -170 = 0**

**(2X^2 -36X + 154 = 0)1/2**

**X^2 – 18X + 77 = 0**

**Factoring X^2 -18X + 77 = 0**

**(X – 7) (X – 11) = 0**

**X – 7 = 0 therefore X = 7**

**X – 11 = 0 therefore X = 7**

**The two number are 7 and 11.**

**Problem Number Two :**

**One leg of a right triangle is 7 cm shorter than the other leg. Its area is 30cm^2. Find its perimeter.**

**Solution :**

**Let X = length of one leg**

**X + 7 = length of the other leg**

**Area of a triangle = (bh)/2**

**Working Equation :**

**X (X+7)/2 = 30**

**(X(X+7)/2 = 30) 2**

**X^2 + 7X = 60**

**X^2 + 7X - 60 = 0**

**Factoring X^2 + 7X -60**

**(X -5 ) ( X + 12 ) = 0**

**X – 5 = 0 therefore X = 5**

**Leg1 = 5**

**Leg2 = 5 + 7 = 12**

**Finding the hypotenuse:**

**Hypotenuse = SQRT(5^2 + 12^2) = SQRT(25 + 144) = SQRT (169) =13**

**Perimeter = 5+12+13 = 30 vcentimeters.**

**Problem Number Three :**

**One side of a triangle is one dm. less than its diagonal. If the perimeter of the rectangle is 62, find the area of the rectangle.**

**Solution :**

**Length of the rectangle = d – 1**

**To represent width use phytagorean theorem and consider diagonal as hypotenuse:**

**d^2 = w^2 + (d – 1)^2**

**Rearranging the equation : w^2 = d^2 - (d-1)^2**

**w^2 = d^2 –d^2 +2d – 1**

**w^2 = 2d -1**

**w = SQRT(2d -1)**

**Working Equation : Based from Perimeter = 2l + 2W**

**2(d-1) + 2 SQRT(2d- 1 ) = 62**

**2d – 2 + 2 SQRT(2d-1) = 62**

**(2SQRT(2d -1 ) = 64 -2d) ½**

**SQRT(2d – 1) = 32 – d**

**(SQRT(2d -1 ))^2 = (32 –d)^2**

**2d -1 = 1024 -64d + d^2**

**d^2 - 66d + 1025 = 0**

**Factoring gives (d – 41) (d -25) = 0**

**d – 25 = 0**

**d = 25**

**Length = d – 1 = 25 – 1 = 24**

**Width = SQRT(2d-1) =SQRT(50- 1) = SQRT49 = 7**

**Area of the rectangle = (24)(7) = 168 dm^2**

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## Comments 14 comments

You must be a really good maths teacher:-)

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OH drat I lost the first comment. The enemy can take a hike to the pits where he is doomed.in the name above all names Jesus Christ.

This really got me to thinking about how our Lord GOD in all His creation came up with the 10 numbers that have a huge huge massive impact in each area of our life. Gods mind is omnipresent He is so amazing.

I think if you were my math teacher i may have understood a whole lot more. I just did not get math. Now today when I study scripture I see Gods creation of numbers in a spiritual manner and how finite my mind really is. This is really a thought provoking hub for me. Have you given thought to Gods number system and how massive it has been in all of time?

I am sure you have you since you are a teacher of math. That is amazing. Well sister so nice to visit you. I do lift you in prayer and hope all is well with you and yours. You keep on in the Light sis. You will I kinow it Phil 4:13. I just love ya, you are precious. Gods Blessings always and 4 ever sista. In Christ.

tnx for this site

ohhhhhhhhh that's cool i can make my assignment now

can you please tell me if what is 5 main problems invoving quadratic equations?

Tnx For the Problems Now i know how toooo

tnx for the problems and solution

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