Solve x2 – 10x + 25 = 35 for x.

1 answer:

3 0

$x^2-10x+25=35\\\\x^2-2\cdot5\cdot x + 5^2=35\ \ \ |\text{use:}\ (a-b)^2=a^2-2ab+b^2\\\\(x-5)^2=35\to x-5=\pm\sqrt{35}\ \ \ |+5\\\\\boxed{x=5-\sqrt{35}\ \vee\ x=5+\sqrt{35}}$

$\text{other method}\\\\x^2-10x+25=35\\\\x^2-5x-5x+25=35\\\\x(x-5)-5(x-5)=35\\\\(x-5)(x-5)=35\\\\(x-5)^2=35\\.....$

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The length of a rectangle is 4 more than it's width. Write an equation that will determine the dimensions if the area of the rec

Firdavs [7]

Answer: The equation is W^2 + 4W - 96= 0

{Please note that ^2 means raised to the power of 2}

Step-by-step explanation: We have been given hints as to the measurement of the length and width of the rectangle. The length is given as four more than the width. What that means is that whatever is the width, we simply add four to get the measurement of the length. Therefore if the width is W, then the length is W + 4.

That is,

L = W + 4 and

W = W

Also we have the area given as 96.

Remember that the area of a rectangle is given as

Area = L x W.

In this question, the Area is expressed as

Area = (W + 4) x W

96 = W^2 + 4W

Subtract 96 from both sides of the equation and we have

W^2 + 4W - 96 = 0.

We now have a quadratic equation from which we can determine the dimensions of the rectangle