Word Problems Using Quadratic Formula

We will discuss here how to solve the word problems using quadratic formula.

We know the roots of the quadratic equation ax$^{2}$ + bx + c = 0, where a ≠ 0 can be obtained by using the quadratic formula x = $\frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.

1. A line segment AB is 8 cm in length. AB is produced to P such that BP$^{2}$ = AB ∙ AP. Find the length of BP.

Solution:

Let BP = x cm. Then AP = AB + BP = (8 + x) cm.

Therefore, BP$^{2}$ = AB ∙ AP

⟹ x$^{2}$ = 8 ∙ (8 + x)

⟹ x$^{2}$ - 8x - 64 = 0

Therefore, x = $\frac{-(-8) \pm \sqrt{(-8)^{2} - 4\cdot 1\cdot (-64)}}{2}$

x = $\frac{-8 \pm \sqrt{64 × 5}}{2}$ = $\frac{-8 \pm 8\sqrt{5}}{2}$

Therefore, x = 4 ± 4√5.

But the length of BP is positive.

So, x = (4 + 4√5) cm = 4(√5 + 1) cm.

2. In the Annual Sports Meet in a girls’ school, the girls present in the meet, when arranged in a solid square has 16 girls less in the front row, than when arranged in a hollow square 4 deep. Find the number of girls present in the Sports Meet.

Solution:

Let the number of girls in the front row when arranged in a hollow square be x.

Therefore, total number of girls = x$^{2}$ - (x - 2 × 4)$^{2}$

= x$^{2}$ - (x - 8)$^{2}$

Now, total number of girls when arranged in Solid Square

= (x - 16)$^{2}$

According to the condition of the problem,

x$^{2}$ - (x - 8)$^{2}$ = (x - 16)$^{2}$

⟹ x$^{2}$ - x$^{2}$ + 16x - 64 = x$^{2}$ - 32x + 256

⟹ -x$^{2}$ + 48x - 320 = 0

⟹ x$^{2}$ - 48x + 320 = 0

⟹ x$^{2}$ - 40x - 8x + 320 = 0

⟹ (x - 40)(x - 8) = 0

x = 40 or, 8

But x = 8 is absurd, because the number of girls in the front row of a hollow square 4 deep, must be greater than 8,

Therefore, x = 40

Number of girl students present in the Sports Meet

= (x - 16)$^{2}$

= (40 - 16)$^{2}$

= 24$^{2}$

= 576

Therefore, the required number of girl students = 576

3. A boat can cover 10 km up the stream and 5 km down the stream in 6 hours. If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.

Solution:

Let the speed of the boat in still water be x km/hour.

Then, the speed of the boat up the stream (or against the stream) = (x - $\frac{3}{2}$) km/hour, and the speed of the boat down the stream (or along the stream) = (x + $\frac{3}{2}$) km/hour.

Therefore, time taken to travel 10 km up the stream = $\frac{10}{x - \frac{3}{2}}$ hours and time taken to travel 5 km down the stream = $\frac{5}{x + \frac{3}{2}}$ hours.

Therefore, from the question,

$\frac{10}{x - \frac{3}{2}}$ + $\frac{5}{x + \frac{3}{2}}$ = 6

⟹ $\frac{20}{2x - 3}$ + $\frac{10}{2x + 3}$ = 6

⟹ $\frac{10}{2x - 3}$ + $\frac{5}{2x + 3}$ = 3

⟹ $\frac{10(2x + 3) + 5(2x – 3)}{(2x – 3)(2x + 3)}$ = 3

⟹ $\frac{30x + 15}{4x^{2} - 9}$ = 3

⟹ $\frac{10x + 5}{4x^{2} - 9}$ = 1

⟹ 10x + 5 = 4x$^{2}$ – 9

⟹ 4x$^{2}$ – 10x – 14 = 0

⟹ 2x$^{2}$ -5x – 7 = 0

⟹ 2x$^{2}$ - 7x + 2x - 7= 0

⟹ x(2x - 7) + 1(2x - 7) = 0