quadratic equation to find area of rectangle

#QUADRATIC EQUATION TO FIND AREA OF RECTANGLE HOW TO#

These examples have shown that a quadratic equation may have two real solutions, one real solution, or two complex solutions. The following example is a little different. Most of the quadratic equations you've looked at have two solutions, like the one above. The power of the Quadratic Formula is that it can be used to solve any quadratic equation, even those where finding number combinations will not work. Sometimes, it may be easier to solve an equation using conventional factoring methods, like finding number pairs that sum to one number (in this example, 4) and that produce a specific product (in this example −5) when multiplied.

#QUADRATIC EQUATION TO FIND AREA OF RECTANGLE HOW TO#

However, upon looking at x 2 + 4 x = 5, you may have thought “I already know how to do this I can rewrite this equation as x 2 + 4 x – 5 = 0, and then factor it as ( x + 5)( x – 1) = 0, so x = −5 or 1.” This is correct-and congratulations if you made this connection! You’ve solved the equation successfully using the Quadratic Formula! You get two true statements, so you know that both solutions work: x = 1 or −5. This means the correct answer is a = 1, b = 3, and c = −6. Putting the terms in order gives the standard form x 2 + 3 x – 6 = 0. You correctly found that 3 x + x 2 = 6 becomes 3 x + x 2 – 6 = 0. Remember that in standard form, the equation is written in the form ax 2 + bx + c = 0. 3 x + x 2 = 6 becomes 3 x + x 2 – 6 = 0, so the standard form is x 2 + 3 x – 6 = 0. This means the correct answer is a = 1, b = 3, and c = −6.Ĭorrect.

The c must be on the left side of the equation. You put the terms in the correct order, but the right side must be equal to 0. If you manage to get through this spoiler successfully I think you will have learned your homework lesson.Incorrect. I do not feel too bad about the plot spoiler here as you have enough work in front of yourself drawing this diagram, setting up the quadratic, and solving it. You will get the same solution by any of the methods. Note you will get two solutions to your quadratic. The width of your path should be $rac$. You could also try taking the top and bottom path areas, then add the remaining unaccounted for height portions, but I will save that for you to see. The two remaining unaccounted for pieces have area $17x$ each, and there are two of these, so solve There are two of these, and so your left and right paths have area $2x(11+2x)$. For example, the outer rectangular height is $11+2x$, and the width is $x$, so the left piece including everything on the path from top to bottom will have area $x(11+2x)$. You could also try an additive process in more than one way. One way would be to take the outer rectangular area, subtract the inner rectangular area, set that all equal to 165, and solve for $x$. There are at least 5 ways to solve this, it just depends upon how you think.

Note the lengths $x$ occur on either end (these are the corners). The height length will be $11+x+x=11+2x$. The outer rectangle will have width length $17+x+x=17+2x$ (don't need length, but make sure your diagram is right). What you should be looking at now is an 11 by 17 inner rectangle. Assign the variable $x$ as being the uniform width on the path between your inner and outer rectangle. Now circumscribe a rectangle around that rectangle of uniform width. Draw a rectangle with width 17 and height 11.

Quadratic equation to find area of rectangle