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Solving quadratic equation

My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!
Reply 1
Original post by miaofcourse
My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!


Your expansion is wrong: (x+3)(3x2)=x(3x2)+3(3x2)=3x22x+9x6=104(x+3)(3x-2) = x(3x-2) + 3(3x-2) = 3x^2 - 2x + 9x - 6 = 104

Do you know that (a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d) = ac + ad + bc + bd?
Original post by miaofcourse
My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!


(x+3)(3x2)=x(3x2)+3(3x2)= ?(x+3)(3x-2) = x(3x-2) + 3(3x-2) =\ ?
Reply 3
Original post by Zacken
Your expansion is wrong: (x+3)(3x2)=x(3x2)+3(3x2)=3x22x+9x6=104(x+3)(3x-2) = x(3x-2) + 3(3x-2) = 3x^2 - 2x + 9x - 6 = 104

Do you know that (a+b)(c+d)=ac+ad+bc+bd(a+b)(c+d) = ac + ad + bc + bd?


Thank you!! I see it now, in my working out I thought it was x+3x not x+3 haha
Reply 4
Original post by miaofcourse
Thank you!! I see it now, in my working out I thought it was x+3x not x+3 haha


Ahhh, glad it's sorted now!
Reply 5
Original post by Zacken
Ahhh, glad it's sorted now!


so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?
Original post by miaofcourse
so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?


When either bracket = 0 the whole thing = 0
and we know x>0
Reply 7
Original post by miaofcourse
so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?


Hey! That's not correct.

You started with (3x2)(x+3)=3x2+7x6=104(3x-2)(x+3) = 3x^2 + 7x - 6 = 104 then make it =0=0 to get 3x2+7x110=03x^2 + 7x -110 = 0 and now solve this by factorising or using the quadratic formula.
Reply 8
Original post by Zacken
Hey! That's not correct.

You started with (3x2)(x+3)=3x2+7x6=104(3x-2)(x+3) = 3x^2 + 7x - 6 = 104 then make it =0=0 to get 3x2+7x110=03x^2 + 7x -110 = 0 and now solve this by factorising or using the quadratic formula.

So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!
Reply 9
Original post by miaofcourse
So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!


looks right
Original post by miaofcourse
So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!


Plug it back through the rectangle lengths to check.

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