Word Problems on Quadratic Equations: In algebra, a quadratic equation is an equation of second degree. If a quadratic polynomial is equated to zero, then we can call it a quadratic equation. The ancient mathematician Sridharacharya derived a formula known as a quadratic formula for solving a quadratic equation by completing the square.

Quadratic equations arise in many situations in the world around us and in the various field of mathematics. There are always two solutions to a quadratic equation. There are multiple methods to solve quadratic equations. In this article let us look at some of the different ways of solving Quadratic Equations Problems using sridharacharya method and more.

Definition of Quadratic Equation

An equation of second-degree polynomial in one variable, such as \(x\) usually equated to zero, is called a quadratic equation. The coefficient of \(x^{2}\) must not be zero in a quadratic equation. If \(p(x)\) is a quadratic polynomial, then \(p(x)=0\) is called a quadratic equation.

Some examples are, \(3 x^{2}+2 x+2=0,-x^{2}+6 x+1=0,7 x^{2}-6 x+4=0\)

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is given by \(a x^{2}+b x+c=0\), where \(a, b, c\) are real numbers, \(a \neq 0\) and \(a\) is the coefficient of \(x^{2}, b\) is the coefficient of \(x\), and \(c\) is a constant.

Quadratic Equation Word Problems

Quadratic equations deal with many real-life situations.

For example, suppose a builder decides to build a community hall of a building having a carpet area of \(1000\,{\text{square}}\,{\text{meter}}\) with its length two meters more than twice its breadth. What should be the length and breadth of the hall?

In such cases, we take the help of quadratic equations to solve and obtain the measures.

Suppose the breadth of the hall is \(x\,{\text{meters}}\). Then, its length will be \((2 x+2)\) \({\text{meters}}\).
Therefore, the area of the hall \( = {\text{length}} \times {\text{breadth}} = x \times (2x + 2) = 1000 \Rightarrow 2{x^2} + 2x = 1000\)

\(\Rightarrow 2 x^{2}+2 x-1000=0\)

Now, this is a quadratic equation.

We can solve any word problems on a quadratic equation using various methods. Let us know about these.

Solving Quadratic Equation Sums Using Factorisation Method

If we can factorize \(a x^{2}+b x+c, a \neq 0\), into a product of two linear factors, then the roots of the quadratic equation \(a x^{2}+b x+c=0\) can be found by equating each factor to zero.
Let us take an example of a word problem.
Example: The product of two consecutive even integers is equal to \(24\). Find these integers.

Let the two integers be \(x, x+2\).

Now, according to the statement, \(x(x+2)=24 \Rightarrow x^{2}+2 x-24=0\)

1. Compare the given quadratic equation with the standard form \(a x^{2}+b x+c=0\) and find the coefficients of \(x^{2}, x\), and the constant to get the values for \(a, b, c\). So, comparing \(x^{2}+2 x-24=0\) with \(a x^{2}+b x+c=0\) we get, \(a=1, b=2, c=24\)
2. Now we will split \(b\) as the sum of two numbers such that the product of these two numbers \(=a \times c=a c\)
   We can factorize a quadratic equation when \(b\) can be split in \(v\) and \(w\) or \(b=v+w\) and \(v, w\) are the factors of \(a \times c=a c\).
   So, \(x^{2}-(6-4) x-24=0\)
   \(\Rightarrow x^{2}-6 x+4 x-24=0\)
3. Find the common factor of the first two and last two terms individually, such as the new two terms have the same common factor.
   So, \(x\) is the common factor of the first two terms \(x^{2}-6 x\). Therefore, \(x(x-6)\). \(4\) is the common factor of the last two terms \(4 x-24\).
   Therefore, \(4(x-6)\).
   So, \(x(x-6)-4(x-6)=0\)
4. If we take out the same common factor, then we get the product of two linear polynomials.
   Now, \((x-6)\) is the common factor in the new two terms.
   So, \((x-6)(x-4)=0\)
5. At last, we will apply the zero product rule to find the solutions. Zero product property says that when \(p \times q=0\), then \(p=0\) or \(q=0\).
   Therefore, \((x-6)=0\), or \((x-4)=0\)
   Hence, the solutions are \(x=6, x=4\).

Solving Quadratic Equations Problems Using Completing Square Method

A quadratic equation can be solved by the method of completing the square. Following are the steps to Solve Quadratic Equation Using Completing the Square Method:

\(a x^{2}+b x+c=0, a \neq 0\)

Let us divide \(a\) from the LHS.
\(\Rightarrow x^{2}+\frac{b}{a} x+\frac{c}{a}=0 \Rightarrow x^{2}+2 \times \frac{b}{2 a} \times x+\frac{c}{a}=0\)

Now, to make the perfect square, we need to add and subtract \(\left(\frac{b}{2 a}\right)^{2}\) from LHS

\(\Rightarrow x^{2}+2 \times \frac{b}{2 a} \times x+\left(\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}=0 \Rightarrow\left(x+\frac{b}{2 a}\right)^{2}-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}=0\)

Transferring \(-\left(\frac{b}{2 a}\right)^{2}+\frac{c}{a}\) from LHS to RHS we have,

\(\Rightarrow\left(x+\frac{b}{2 a}\right)^{2}=-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}\)

Taking square root on both sides we have,

\(\Rightarrow x+\frac{b}{2 a}=\pm \sqrt{-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}}\)
\(\Rightarrow x=\pm \sqrt{-\frac{c}{a}+\left(\frac{b}{2 a}\right)^{2}}-\frac{b}{2 a}\)

Example: A motorboat whose speed is \(18\,{\text{kmph}}\) in still water takes \(1\,{\text{hour}}\) more to go \(24 \mathrm{~km}\) upstream than to return downstream to the exact location. Find the speed of the stream.

Let the speed of the stream be \(x\,{\text{kmph}}\).

We know that when the boat travels upstream, the relative speed between the boat and the stream \(=(18-x)\,{\text{kmph}}\)

When the boat travels downstream, the relative speed between the boat and the stream \(=(18+x)\,{\text{kmph}}\)

The time required to go upstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18-x}\) \({\text{hours}}\)

Similarly, the time required to go downstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{24}{18+x}\) \({\text{hours}}\)

Now, according to the question, \(\frac{24}{18-x}-\frac{24}{18+x}=1 \Rightarrow 24\left[\frac{1}{18-x}-\frac{1}{18+x}\right]=1 \Rightarrow\left[\frac{1}{18-x}-\frac{1}{18+x}\right]=\frac{1}{24}\)

To solve the for \(x\) we need to convert the above equation into the standard form of the quadratic equation.
\(\Rightarrow\left[\frac{(18+x)-(18-x)}{(18-x)(18+x)}\right]=\frac{1}{24} \Rightarrow\left[\frac{18+x-18+x}{(18-x)(18+x)}\right]=\frac{1}{24}\)
\(\Rightarrow\left[\frac{2 x}{18^{2}-x^{2}}\right]=\frac{1}{24}\) (using the algebraic identity \(a^{2}-b^{2}=(a+b)(a-b)\)

Now, cross multiplying we have, \(\Rightarrow 48 x=324-x^{2}\)
\(\Rightarrow x^{2}+48 x-324=0 \Rightarrow x^{2}+2 \times 24 \times x-324=0\)

Now, to make the perfect square, we need to add and subtract \((24)^{2}\) from LHS
\(\Rightarrow x^{2}+2 \times 24 \times x+(24)^{2}-(24)^{2}-324=0 \Rightarrow(x+24)^{2}-(24)^{2}-324=0\)

Transferring \(-(24)^{2}-324\) from LHS to RHS we have,

\(\Rightarrow(x+24)^{2}=+324+(24)^{2}\)

Taking square root on both sides we have,
\(\Rightarrow x+24=\pm \sqrt{900}\)
\(\Rightarrow x=\pm 30-24\)
\(\Rightarrow x=+30-24\)
\(\Rightarrow x=-30-24=-54\)

We will consider the positive value of \(x\) as speed can not be negative.

Hence, the speed of the stream is \(6\,{\text{kmph}}\).

Solving Quadratic Equations Problems Using Sridharacharya Method

The roots of the quadratic equation \(a x^{2}+b x+c=0\) are given by the quadratic formula
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)

Example: Madhu and Rimi have \(45\) marbles. Both of them lost \(5\) marbles each, and the product of the marbles they have now is \(124\). Find out how many marbles they had in the beginning?

Let us assume the number of marble Madhu had is \(x\) and the number of marbles Rimi had is \((45-x)\).

The number of marbles Madhu had when she lost \(5\) marbles \(=x-5\)

The number of marbles Rimi had when she lost \(5\) marbles \(=45-x-5=40-x\)

According to the statement
\(\left({x – 5} \right)\left({40 – x} \right) = 124 \Rightarrow  – {x^2} + 40x + 5x – 200 = 0\)
\( \Rightarrow  – {x^2} + 45x – 200 – 124 = 0 \Rightarrow {x^2} – 45x + 324 = 0\)

From the given quadratic equation \(a=1, b=-45, c=324\). So, by the quadratic formula
\(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
\(=\frac{-(-45) \pm \sqrt{(-45)^{2}-4 \times 1 \times 324}}{2 \times 1}=\frac{+45 \pm \sqrt{2025-1296}}{2}\)
\(=\frac{45 \pm \sqrt{729}}{2}=\frac{45 \pm 27}{2}=\frac{45+27}{2}, \frac{45-27}{2}=36,9\)
\(\Rightarrow x=36\) or \(x=9\)

Hence, they had started with \(36\) and \(9\) marbles.

If Madhu had \(36\) marbles, then Rimi had \(9\) marbles, and if Madhu had \(9\) marbles, then Rimi had \(36\) marbles.

Solved Examples – Problems on Quadratic Equations for Class 10

Q.1. A boat has a speed of \(16 \mathrm{~km} / \mathrm{hr}\) in still water. In a particular river, it takes \(2\) hours more to go \(32 \mathrm{~km}\) upstream than to return to the same point. What is the speed at which water is flowing in the river?
Ans: Let the speed of the stream be \(x\,{\text{kmph}}\).
We know that when the boat travels upstream, the relative speed between the boat and the stream \((16-x)\,{\text{kmph}}\)
When the boat travels downstream, the relative speed between the boat and the stream \((16+x)\,{\text{kmph}}\)
The time required to go upstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{32}{16-x}\) \({\text{hours}}\)
Similarly, the time required to go downstream \(=\frac{\text { Distance }}{\text { Speed }}=\frac{32}{16+x}\) \({\text{hours}}\)
Now, according to the question,
\(\frac{32}{16-x}-\frac{32}{16+x}=2 \Rightarrow 32\left[\frac{1}{16-x}-\frac{1}{16+x}\right]=2 \Rightarrow\left[\frac{1}{16-x}-\frac{1}{16+x}\right]=\frac{2}{32}\)
To solve the for \(x\) we need to convert the above equation into the standard form of the quadratic equation.
\(\Rightarrow\left[\frac{(16+x)-(16-x)}{(16-x)(16+x)}\right]=\frac{1}{16} \Rightarrow\left[\frac{16+x-16+x}{(16-x)(16+x)}\right]=\frac{1}{16}\)
\(\Rightarrow\left[\frac{2 x}{16^{2}-x^{2}}\right]=\frac{1}{16}\) (using the algebraic identity \(a^{2}-b^{2}=(a+b)(a-b)\)
Now, cross multiplying we have, \(\Rightarrow 32 x=256-x^{2}\)
We have \(\Rightarrow x^{2}+32 x-256=0\)
\(\Rightarrow x^{2}+2 \times 16 \times x-256=0\)
Now, to make the perfect square, we need to add and subtract \((16)^{2}\) from LHS \(\Rightarrow x^{2}+2 \times 16 \times x+(16)^{2}-(16)^{2}-256=0 \Rightarrow(x+16)^{2}-(16)^{2}-256=0\)
Transferring \(-(16)^{2}-256\) from LHS to RHS we have,
\(\Rightarrow(x+16)^{2}=512\)
Taking square root on both sides we have, \(\Rightarrow x+16=\pm \sqrt{512}\)
\(\Rightarrow x=\pm 16 \sqrt{2}-16\)
\(\Rightarrow x=+16 \sqrt{2}-16\) or, \(x=-16 \sqrt{2}-16\)
We will consider the positive value of \(x\) as speed can not be negative. Hence, the speed at which the water is flowing in the river is \( + 16\sqrt 2  – 16\,{\text{kmph}}\)

Q.2. Priya decides to build a prayer hall of her house having a carpet area of \(300\,{\text{square}}\,{\text{meters}}\) with its length of \(25\,{\text{meters}}\) more than its breadth. What should be the breadth and length of the prayer hall?
Ans: Let us assume breadth is \(x\), and the length is \((x+25)\). According to the question,
\(x(x+25)=300 \Rightarrow x^{2}+25 x-300=0\)
\(\Rightarrow\) From the given quadratic equation \(a=1, b=25, c=-300\)
Quadratic equation formula is given by \(x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
\(x=\frac{-(25) \pm \sqrt{(25)^{2}-4 \times 1 \times(-300)}}{2 \times 1}=\frac{-25 \pm \sqrt{1825}}{2}\)
\(x=\frac{-25+\sqrt{1825}}{2}=8.86\) (approx) or \(x=\frac{-25-\sqrt{1825}}{2}=-33.86\) (approx) We will avoid the negative value of \(x\) as breadth can not be negative.
Hence, the breadth is, \(x=8.86\) (approx) and the length is \(x+25=8.86\) (approx) \(+25=33.86\) (approx)

Q.3. The product of two odd consecutive positive integers is \(63\). Find the numbers.
Ans: Let the two odd integers be \(x, x+2\). We have:
Now, according to the statement,
\(x\left({x + 2} \right) = 63 \Rightarrow {x^2} + 2x – 63 = 0\)
\( \Rightarrow {x^2} + \left({9 – 7} \right)x – 63 = 0 \Rightarrow x^{2} + 9x – 7x – 63 = 0\)
\( \Rightarrow x\left({x + 9} \right) – 7\left({x + 9} \right) = 0 \Rightarrow \left({x + 9} \right)\left({x – 7} \right) = 0\)
So, \(x=-9\) or, \(x=7\)
Here, we need to find the positive odd integers. So, avoid the negative value of \(x\).
Hence, the two odd consecutive integers are \(x=7\) and \(x+2=7+2=9\).

Q.4. The product of Rime’s age two years ago and her age four years from now is one more than twice her present age. What is her present age?
Ans: Let her present age be \(x\).
Two years ago, her age was \((x-2)\).
Her age after four years from now is \((x+4)\).
One more than twice her present age is \((1+2 x)\).
According to the statement,
\((x-2)(x+4)=1+2 x x^{2}+4 x-2 x-8=1+2 x\)
\(x^{2}+2 x-2 x-8-1=0\)
\(x^{2}-9=0 x^{2}=9\)
Thus, \(x=\pm 3\)
Now, age can not be negative.
Hence, Rime’s present age is \(3\) years

Q.5. If the speed of a car is increased by \(10\,{\text{km}}/{\text{hr}}\), the time of journey for a distance of \(72\,{\text{km}}\) is reduced by \(36\,{\text{minutes}}\). Find the initial speed of the car.
Ans: Let the initial speed of the car is \(x\,{\text{kmph}}\).
Speed of the car after increasing the speed is \((x+10),{\text{kmph}}\).
Distance is \(72 \mathrm{~km}\) (given).
Time taken to cover the distance is \(\left( {\frac{{72}}{x}} \right)\,{\text{hours}}\)
Time taken after increasing the speed is \(\left( {\frac{{72}}{x+10}} \right)\,{\text{hours}}\)
According to the question,
\(\Rightarrow \left( {\frac{1}{x}} \right) – \left( {\frac{1}{{x + 10}}} \right) = \frac{{36}}{{60 \times 72}}\)
\(\Rightarrow \left[{\frac{{x + 10 – x}}{{x\left({x + 10} \right)}}} \right] = \frac{1}{{120}} \Rightarrow {x^2} + 10x – 1200 = 0\)
\( \Rightarrow \,\,{x^2} + 40x – 30x – 1200 = 0\)
\( \Rightarrow \,\,x\left({x + 40} \right) – 30\left({x + 40} \right) = 0\)
\(\Rightarrow \left({x + 40} \right)\left({x – 30} \right) = 0 \Rightarrow x =  – 40\) or \(x = 30\)

Summary

In this article, we discussed quadratic equation in the variable \(x\), which is an equation of the form \(a x^{2}+b x+c=0\), where \(a, b, c\) are real numbers, \(a \neq 0\). Also, we discussed the methods of solving the quadratic equations, such as factorizing method, completing the square method, quadratic formula method, etc.

FAQs on Word Problems on Quadratic Equation

We have given the answers to the most commonly asked questions on Word Problems on Quadratic Equations:

Q.1. What are the three types of quadratic equations? Ans: There are three types of quadratic equations 1. Standard form: \(a x^{2}+b x+c=0, a \neq 0\) 2. Factored form: \((x-a)(x-b)=0\) 3. Vertex form: \(a(x-h)^{2}+k=0\) Each form of a quadratic equation has unique importance.

Q.2. How do you write a quadratic equation from a word problem? Ans: The form \(a x^{2}+b x+c=0, a \neq 0\) is called the standard form of a quadratic equation. According to the statement, we will consider the unknown value as \(x\). Then we will make the equation using the given values in the question. After that, we will convert it into the standard form of the quadratic equation. Finally, we will solve for \(x\) using one of the methods of solving quadratic equation which is relevant to the given word problem.

Q.3. Are all given quadratic equations in standard form? Ans: No, all given quadratic equations may not be in the standard form. To solve for the unknown values, we need to convert the given equation into the standard form. The form \(a x^{2}+b x+c=0, a \neq 0\) is called the standard form of a quadratic equation.

Q.4. What are the \(5\) examples of a quadratic equation? Ans: The problem related to (i) Throwing a ball (ii) A parabolic mirror (iii) Shooting a cannon (iv) Diving from a platform (v) Hitting a golf ball All these can be solved using the quadratic equation.

Q.5. What types of problems can you solve using quadratic equations? Ans: Quadratic equations deal with many real-life situations. For example, suppose a builder decides to build a community hall of a building having a carpet area of \(1000\,{\text{square}}\,{\text{meter}}\) with its length of two meters more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is \(x\) meters. Then, its length will be \((2 x+2)\) \({\text{meter}}\). Therefore, the area of the hall \(\, = {\text{length}} \times {\text{breadth}} = x \times (2x + 2) = 1000 \Rightarrow 2{x^2} + 2x = 1000 \Rightarrow 2{x^2} + 2x – 1000 = 0\)

We hope you find this detailed article on word problems on quadratic equations helpful. We expect you to solve more questions on our platform on a wide range of topics including quadratic equations. This will build your confidence in the topic.