Quadratic Equations Test - 2

The smaller of the two consecutive positive odd integers is x.
Then, the second odd integer is x + 2.
According to the question:
x (x + 2) = 63
⇒ x² + 2x = 63 or x² + 2x – 63 = 0 is the required quadratic equation.

(2y – 5)/(3y – 7) = (y – 1)/(y + 1)
(2y – 5)(y + 1) = (y – 1)(3y – 7)
2y² – 3y - 5 = 3y² - 10y + 7
y² – 7y + 12 = 0
y² – 3y – 4y + 12 = 0
(y – 3)(y – 4) = 0
y = 3, 4

Let the original number be x.
Now, as per given information:
(x + 16)² = 2x² + 368
x² + 256 + 32x = 2x² + 368
x² - 32x + 112 = 0
(x - 28)(x - 4) = 0
x = 28 or 4

Let the side of the smaller square be x cm.
Then, length of each side of the larger square = (x + 4) cm
According to problem:
(x + 4)² + x² = 400 {∵ Area of square = Side × Side}
x² + 8x + 16 + x² = 400
2x² + 8x - 384 = 0
x² + 4x - 192 = 0
(x + 16)(x - 12) = 0
x = 12, x ≠ -16
So, length of side of the larger square = (12 + 4) cm = 16 cm
Area = (16)² cm² = 256 cm²

(1) x² + x - 5 = 0
D = 1 - 4(1)(-5) = 1 + 20 = 21 > 0
This equation has real roots.

(2) 5x² - 3x + 1 = 0
D = (-3)² - 4(5)(1) = 9 - 20 = -11 < 0
This equation does not have real roots

(3) -x² + 2x + 3 = 0
D = (2)² - 4(-1)(3) = 4 + 12 = 16 > 0
This equation has real roots.

(4) x² + 3x - 12 = 0
D = (3)² - 4(-12)(1) = 9 + 48 = 57 > 0
This equation has real roots.