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.
Algebra/Linear Algebra
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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