I need help solving this word problem relating to quadratic equations?

inverted com(m)as = quotation marks

let a = side of a square and b = side of another square, a and b both greater than 0

"The perimeter of a square exceeds that of another by 100cm"

The perimeter of a square is 4 times the length of a side

4a = 4b+100

"the area of the larger square exceeds three times the area of the smaller square by 325cm sq

The area of a square is the square of the length of a side

a² = 3b² + 325

You have two independent equations in two unknowns so you can solve them for the values of the unknowns.

4a = 4b+100 → a = b+25

a² = 3b² + 325 → (b+25)² = 3b² + 325

b² + 50b + 625 = 3b² + 325

2b² - 50b - 300 = 0

b² - 25b - 150 = 0

(b-30)(b+5) = 0

b = -5, 30 ← discard the negative value since b > 0

b = 30

Now that we know b, we can use the equation a = b+25 to find a

a = 30+25 = 55

We now know that the length of the side of the smaller square is 30cm and the length of the larger square is 55cm.

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Check:

Perimeter:

Pa = 4(55) = 220cm

Pb = 4(30) = 120cm

difference = 100cm ← Checks

Area:

Aa = 55² = 3025cm²

Ab = 30² = 900cm²

3025 - 3*900 = 325cm² ← Checks

Let, a and b be the sides of the two squares.

4b = 4a + 100

this implies that b = a + 25

b^2 = 3a^2 + 325

Substituting value of b in this 2nd equation, we get,

(a + 25)^2 = 3a^2 + 325

a^2 + 625 + 50a = 3a^2 + 325

2a^2 - 50a - 300 = 0

a^2 -25a - 150 = 0

a = {25 +- sqrt[25^2 - 4(1)(-150)]}/2

a = {25 +- 35}/2

a = 30, -5

but a can't be -ve, so a = 30

therefore, b = a + 25 = 30 + 25 = 55