Quadratic equations

[bumblebee123]

I've been working on this question for a long time, but I still can't get the right answer. can someone tell me what I'm doing wrong?

Question: a square has the sides of length ( x + 2 ) cm
a right-angled isosceles triangle has its two equal sides of length ( 2x + 1 ) cm
the area of the square is equal to the area of the triangle

a) write down an equation in x

I worked this out to be 1/2 ( 2x+ 1 ) ^2 = ( x + 2 ) ^2 which is the correct answer

b) show that your equation simplifies to 2x^2 - 4x - 7 = 0 I managed to do this by just expanding the brackets from the previous question

c) by solving the equation 2x^2 - 4x - 7 = 0, find the perimeter of the square ( give your answer to 3 s.f. )

This is where I got stuck. To find the value of x ( to plug into the equation for the side of the square ) I put the equation into the quadratic formula:

x = ( 4 +- sqrt 72 ) / 4

this gave me: x = 3.12 and x= -1.12

by plugging the x values into the formula for the side of the square ( x + 2 ) and then multiplying by 4 to get the perimeter I got the wrong answer ( 20 ).

The correct answer is 12.5 cm

any help would be really appreciated?

 

[Dr.Peterson]

It looks like you did it correctly (except for a rounding issue), while they made the mistake I would expect a student to make: they forgot that the side is x+2, and just multiplied x = 3.12 by 4 to get 12.48, rounded to 12.5.

What should the answer really be (rounded to 3 sf)?

 

[bumblebee123]

( 3.121320344 + 2 ) x 4 = 5.122320344 x 4 = 20.48528137 cm = 20.5 cm

does this mean that the answer in the book is wrong?

 

[Dr.Peterson]

Yes, that's exactly what I said. They made a silly mistake. That happens.