- QUADRATIC EQUATION TO FIND AREA OF RECTANGLE HOW TO
- QUADRATIC EQUATION TO FIND AREA OF RECTANGLE PLUS
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. And then you can verify.Ok what you want to do here is this. The base of a triangle can't have length of negative 6. Whether this makes sense in the context of We can't just state, oh b couldīe negative 6 or b could be 10. The solutions are either b is equal to negativeĦ, or b is equal to 10.
QUADRATIC EQUATION TO FIND AREA OF RECTANGLE PLUS
Out a b plus 6 here, you get 0 is equal to b The second one you canįactor out a negative 10. Up this negative 4b into its constituents. Sides of this equation, you get b is equal to 10. Sides of this equation, we get b is equal to negative 6. That either one of these is equal to zero. Y is equal to negative 4, and x times y isĮqual to negative 60. That add up to this second term right over here. Here to say, look, we're looking for two numbers To make it very clear, is different than the b that Say that this is equal to b plus 6, timesī, minus 10. So if we make itĦ and negative 10 their sum will be negativeĤ, and their product is negative 60. The larger absolute magnitude number to be negative so Still seems too far apart from each other. Of them negative, their sum would be either
One of the negative, you would either get positiveĥ9 as the sum or negative 59 as the sum. That one is going to beįour less than the others. Their absolute values are going to be four apart. Want to find two numbers whose sum is negative 4 and Means that either one or both of those things need The product of some things, and that equals 0, that To do here is just factor this thing right So let's subtract 60 fromīoth sides of this equation. On one side of the equation, having them equal 0. Term right here- is to get all of the terms Solve a quadratic- we have a second degree So let's multiplyĢ times b squared over 2 is just b squared. Of this fraction here let's multiply both One half times- let's distribute the b- timesī, let me make it clear. Is the same thing as 4 less than the base. Now instead of puttingĪn h in for height, we know that the height Or we could say thatģ0 inches squared is equal to one half times The base times the height we'll get 30 inches squared. Length of this bottom side, that's the base, So let's think aboutĭraw a triangle here. Use the formula areaĮquals one half base times height for theĪrea of a triangle. Is four inches less than the length of the base.
(2x²+8x)+(5x+20)=0 ( I grouped them apropriately so that I can factor out)Ģx(x+4)+5(x+4)=0 ( I factored out a 2 and 5) 8 and 5 because the satisfy all the conditions we are looking for. Multiply them they equal = A*C =q*w= 2*20=40
So we want to find two coefficients of x (q and w) to replace B which when you. Lets say we want to factor an equation the form of Ax²+Bx+C=0 Okay lets have an example so that we can understand it even more. You'll further simplify it with real numbers.
(note:coefficient of means the number beside)Ĭ=the constant term(or the term that doesn't have an x beside it)Īnd we want to find two coefficients of x (q and w) to replace B which when you. The quadratic equation is in the form of.
QUADRATIC EQUATION TO FIND AREA OF RECTANGLE HOW TO
I'll explain how to do factoring by grouping but first be aware that.
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