Introduction: How do you solve quadratic equation word problems
In algebra, a quadratic equation is a polynomial equation of the second degree. The general form is
ax² +bx+c=0
Where x stand for a variable, and a, b, and c, constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)
The term "quadratic" comes from quadratics; Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.
How to Solve Word Problems:
We can solve the word problems by using quadratic equation that is write the problems in quadratic form and find out the result by factorization method or using quadratic formula. In math, the word problems are used to test the student's thinking knowledge. Some example problems for quadratic equation word problems. some times the quadratic formual is used to sovle the word problems.
Here we will explain how do you solve quadratic equation word problems
Example1:
How do you solve the width of a rectangle is 4 inches more than its length. The area of the rectangle is 32 square inches. Find the dimensions of the rectangle.
Solution :
The formula for the rectangle area is LW = A
Step 1 - Draw the picture of the Rectangle.
Since the width is 4more than the length we let the rectangle length = x and the width = x + 4
Step 2 - Write the equation using the x(x +4) = 32
Step 3 - Solve the equation x2 +4x = 32
x2 + 4x.-32= 0
(x+8)(x-4)=0
X+8=0 | x-4=0
X=8 | x=4 the negative value is not established .so, we take x=8
The length of the rectangle is 8inch and width is 12 inch.
Examples
Example2:
How do you solve the width of a rectangle is 6 inches more than its length. The area of the rectangle is 72 square inches. Find the dimensions of the rectangle.Having problem with how to solve a system of equations keep reading my upcoming posts, i will try to help you.
Solution :
The formula for the area of rectangle = LW = A
Step 1 - Draw the picture of the rectangle
Since the width is 4more than the length we let the rectangle length = x and the width = x + 6
Step 2 - Write the equation using the x(x +6) = 72
Step 3 - Solve the equation x2 +6x = 72
x2 + 6x.-72= 0
x2 +12x - 6x -72=0
(x-6)(x+12)=0
X-6=0 | x+12=0
X=6 | x=-12 the negative value is not accepted .so, we take x=6
The length of the rectangle is 6inch and width is 12 inch.
Example 3: The rectangle has 4 inches more than its width and the rectangle area is 90. Find out the length and width of rectangle.
Solution:
Step 1:
The dimensions of rectangle are width = x and length = x + 4.
Step 2:
Rectangle area formula is lw.
A = x(x+4) = 90
Step 3:
The quadratic equation is x2 + 4x – 32 = 0.
Solve the equation,
X2 - 4x + 8x – 32 = 0
X(x – 4) + 8(x – 4) = 0
(x – 4) (x + 8) = 0
X = 4, x = -8.
Therefore, the width of rectangle is 4 and the length of 4 and the width is 8.
Example 4 : Find two positive consecutive odd integers whose product is 63.
Solution:
Let x be the first integer. Then the next odd integer is x + 2. So we have
x(x+2) = 63.
To solve this equation first we distribute and then set one side to zero. We have
x2+2x =63
x2+2x - 63 = 0
Factoring the quadratic equation x2 + 9x - 7x - 63 = 0
x(x + 9) - 7(x + 9) =0
(x + 9)(x - 7) = 0.
So the solutions are x = -9 or x = 7 Since we are looking for positive integers, we take x = 7
Hence the numbers are 7 and 9.
Example 5 :The triangle has three legs. The first triangle leg is 10 inches when the hypotenuse is 12 inches. Find out another leg.
Solution:
Take the unknown leg as x and use the Pythagorean Theorem.
x2 + 102 = 122
Simplify the equation,
x2 + 100 = 144
x2 = 144 – 100
x2 = 44
x = `sqrt(44)`
x = 6.63
Therefore, another leg measurement is 6.63 inches.
The quadratic equations are mainly used in real-time problems.
In algebra, a quadratic equation is a polynomial equation of the second degree. The general form is
ax² +bx+c=0
Where x stand for a variable, and a, b, and c, constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)
The term "quadratic" comes from quadratics; Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.
How to Solve Word Problems:
We can solve the word problems by using quadratic equation that is write the problems in quadratic form and find out the result by factorization method or using quadratic formula. In math, the word problems are used to test the student's thinking knowledge. Some example problems for quadratic equation word problems. some times the quadratic formual is used to sovle the word problems.
Here we will explain how do you solve quadratic equation word problems
Example1:
How do you solve the width of a rectangle is 4 inches more than its length. The area of the rectangle is 32 square inches. Find the dimensions of the rectangle.
Solution :
The formula for the rectangle area is LW = A
Step 1 - Draw the picture of the Rectangle.
Since the width is 4more than the length we let the rectangle length = x and the width = x + 4
Step 2 - Write the equation using the x(x +4) = 32
Step 3 - Solve the equation x2 +4x = 32
x2 + 4x.-32= 0
(x+8)(x-4)=0
X+8=0 | x-4=0
X=8 | x=4 the negative value is not established .so, we take x=8
The length of the rectangle is 8inch and width is 12 inch.
Examples
Example2:
How do you solve the width of a rectangle is 6 inches more than its length. The area of the rectangle is 72 square inches. Find the dimensions of the rectangle.Having problem with how to solve a system of equations keep reading my upcoming posts, i will try to help you.
Solution :
The formula for the area of rectangle = LW = A
Step 1 - Draw the picture of the rectangle
Since the width is 4more than the length we let the rectangle length = x and the width = x + 6
Step 2 - Write the equation using the x(x +6) = 72
Step 3 - Solve the equation x2 +6x = 72
x2 + 6x.-72= 0
x2 +12x - 6x -72=0
(x-6)(x+12)=0
X-6=0 | x+12=0
X=6 | x=-12 the negative value is not accepted .so, we take x=6
The length of the rectangle is 6inch and width is 12 inch.
Example 3: The rectangle has 4 inches more than its width and the rectangle area is 90. Find out the length and width of rectangle.
Solution:
Step 1:
The dimensions of rectangle are width = x and length = x + 4.
Step 2:
Rectangle area formula is lw.
A = x(x+4) = 90
Step 3:
The quadratic equation is x2 + 4x – 32 = 0.
Solve the equation,
X2 - 4x + 8x – 32 = 0
X(x – 4) + 8(x – 4) = 0
(x – 4) (x + 8) = 0
X = 4, x = -8.
Therefore, the width of rectangle is 4 and the length of 4 and the width is 8.
Example 4 : Find two positive consecutive odd integers whose product is 63.
Solution:
Let x be the first integer. Then the next odd integer is x + 2. So we have
x(x+2) = 63.
To solve this equation first we distribute and then set one side to zero. We have
x2+2x =63
x2+2x - 63 = 0
Factoring the quadratic equation x2 + 9x - 7x - 63 = 0
x(x + 9) - 7(x + 9) =0
(x + 9)(x - 7) = 0.
So the solutions are x = -9 or x = 7 Since we are looking for positive integers, we take x = 7
Hence the numbers are 7 and 9.
Example 5 :The triangle has three legs. The first triangle leg is 10 inches when the hypotenuse is 12 inches. Find out another leg.
Solution:
Take the unknown leg as x and use the Pythagorean Theorem.
x2 + 102 = 122
Simplify the equation,
x2 + 100 = 144
x2 = 144 – 100
x2 = 44
x = `sqrt(44)`
x = 6.63
Therefore, another leg measurement is 6.63 inches.
The quadratic equations are mainly used in real-time problems.
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