Width of a rectangular frame in terms of x.

[mathlearn]

By bending a 16 cm wire a rectangular frame is made.

By taking the length as x , write the width in terms of x.

If the area of the frame is $11 cm^2$ show that x satisfies the equation $x^2-8x+11=0$

Solve the equation by taking the $\sqrt{5}=2.24$

Ideas on how to begin ? (Happy)

 

[Greg]

Well, the perimeter (P) is 16 centimetres, and P = 2L + 2W, where L is length and W is width. Can you work with that?

 

[mathlearn]

Well, the perimeter (P) is 16 centimetres, and P = 2L + 2W, where L is length and W is width. Can you work with that?

16=2x+2width

16-2x=2 width

8-x= width

Correct ? :)

If the area of the frame is $11 cm^2$ show that x satisfies the equation $x^2-8x+11=0$

Now area =$11 cm^2$

x*$\left(8-x\right)$=11
$-x^2+8x=11$
$-x^2+8x-11=0$

Correct ? :)

 

[Greg]

Yes! :)

 

[mathlearn]

Solve the equation by taking the $\sqrt{5}=2.24$

Now using the quadratic formula, on $-x^2+8x-11=0$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

$x=\frac{-8\pm\sqrt{64-44}}{-2}$.

$x=\frac{-8\pm\sqrt{20}}{-2}$.

$x=\frac{-8\pm\sqrt{4}\sqrt{5}}{-2}$.

$x={+4\pm-1*\sqrt{5}}$.

$x={+4\pm-1*2.24}$

$x=1.76$. or $x={+4\pm-1*-2.24}=6.76$

Correct? :)