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

⟹ (2x - 7)(x + 1) = 0

⟹ 2x - 7 = 0 or x + 1 = 0

⟹ x = \(\frac{7}{2}\) or x = -1

But speed cannot be negative. So, x = \(\frac{7}{2}\) = 3.5

Therefore, the speed of the board in still water is 3.5 km/h.

**Quadratic Equation**

**Introduction to Quadratic Equation**

**Formation of Quadratic Equation in One Variable**

**General Properties of Quadratic Equation**

**Methods of Solving Quadratic Equations**

**Examine the Roots of a Quadratic Equation**

**Problems on Quadratic Equations**

**Quadratic Equations by Factoring**

**Word Problems Using Quadratic Formula**

**Examples on Quadratic Equations **

**Word Problems on Quadratic Equations by Factoring**

**Worksheet on Formation of Quadratic Equation in One Variable**

**Worksheet on Quadratic Formula**

**Worksheet on Nature of the Roots of a Quadratic Equation**

**Worksheet on Word Problems on Quadratic Equations by Factoring**

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