Chapter 4: Quadratic Equations

We have to solve the quadratic equation

$$2x^2 - 5x + 3 = 0.$$

Step 1 : Split the middle term

Compute the product of the coefficient of $$x^2$$ and the constant term:

$$2 \times 3 = 6.$$

We need two numbers whose

• product is $$6$$, and
• sum is the middle-term coefficient $$-5$$.

The numbers $$-2$$ and $$-3$$ satisfy these conditions, because

$$(-2) + (-3) = -5 \quad\text{and}\quad (-2)\times(-3) = 6.$$

Step 2 : Rewrite the quadratic using the split terms

Replace $$-5x$$ by $$-2x - 3x$$:

$$2x^2 - 2x - 3x + 3 = 0.$$

Step 3 : Group the terms

$$\bigl(2x^2 - 2x\bigr) + \bigl(-3x + 3\bigr) = 0.$$

Step 4 : Factor each group

From the first group, factor $$2x$$; from the second, factor $$-3$$:

$$2x(x - 1) - 3(x - 1) = 0.$$

Step 5 : Factor the common binomial

Both terms contain the factor $$(x - 1)$$, so

$$(x - 1)(2x - 3) = 0.$$

Step 6 : Apply the zero-product property

For a product to be zero, at least one factor must be zero:

$$x - 1 = 0 \quad\text{or}\quad 2x - 3 = 0.$$

Solving each equation, we get

$$x = 1 \quad\text{or}\quad x = \frac{3}{2}.$$

Hence the roots of the quadratic equation are $$x = 1$$ and $$x = \dfrac{3}{2}$$.

The given quadratic equation is $$6x^2 - x - 2 = 0$$.

Compare with the standard form $$ax^2 + bx + c = 0$$ to identify the coefficients:

• $$a = 6$$
• $$b = -1$$
• $$c = -2$$

According to the quadratic formula, the roots are given by $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.

First find the discriminant $$D = b^2 - 4ac$$:

$$D = (-1)^2 - 4 \times 6 \times (-2) = 1 + 48 = 49$$.

Since $$D = 49$$ is a perfect square and positive, the equation has two distinct real and rational roots.

Substitute $$a, b, c$$ and $$D$$ into the quadratic formula:

$$x = \dfrac{-(-1) \pm \sqrt{49}}{2 \times 6} = \dfrac{1 \pm 7}{12}$$.

Taking the “+” sign:

$$x = \dfrac{1 + 7}{12} = \dfrac{8}{12} = \dfrac{2}{3}$$.

Taking the “−” sign:

$$x = \dfrac{1 - 7}{12} = \dfrac{-6}{12} = -\dfrac{1}{2}$$.

Therefore, the roots of the equation $$6x^2 - x - 2 = 0$$ are $$x = \dfrac{2}{3}$$ and $$x = -\dfrac{1}{2}$$.

The given quadratic equation is $$3x^2-2\sqrt{6}\,x+2=0$$.

Compare with the standard form $$ax^2+bx+c=0$$ to obtain

• $$a = 3$$
• $$b = -2\sqrt{6}$$
• $$c = 2$$

The quadratic-formula gives the roots:

$$x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}.$$

Compute the discriminant $$\Delta=b^2-4ac$$:

$$\Delta = \bigl(-2\sqrt{6}\bigr)^2 - 4\times3\times2$$

$$\Delta = 4\times6 - 24 = 24 - 24 = 0.$$

Because $$\Delta=0$$, the equation has real and equal roots. Substitute $$\Delta=0$$ in the formula:

$$x = \dfrac{-b}{2a} = \dfrac{-\bigl(-2\sqrt{6}\bigr)}{2\times3} = \dfrac{2\sqrt{6}}{6} = \dfrac{\sqrt{6}}{3}.$$

Thus both roots are

$$x_1 = x_2 = \dfrac{\sqrt{6}}{3}.$$

Let the breadth (width) of the rectangular prayer hall be $$x\text{ m}$$.

From the information provided in Section 4.1, its length is one metre more than twice the breadth; therefore

$$\text{length} = 2x + 1\text{ m}. $$

The floor area of the hall is given to be $$528\text{ m}^2$$. Hence

$$\text{area} = (\text{length})(\text{breadth}) \;\Rightarrow\; (2x + 1)(x) = 528.$$

Simplifying,

$$2x^2 + x - 528 = 0.$$

This is a quadratic equation in $$x$$. Its discriminant is

$$\Delta = b^2 - 4ac = 1^2 - 4(2)(-528) = 1 + 4224 = 4225.$$

Since $$\Delta > 0$$, the equation has two real roots. Using the quadratic formula,

$$x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-1 \pm \sqrt{4225}}{2\times 2} = \frac{-1 \pm 65}{4}.$$

This gives the values

• $$x = \frac{64}{4} = 16,$$
• $$x = \frac{-66}{4} = -16.5\quad(\text{rejected, length cannot be negative}).$$

Thus, $$x = 16\text{ m}$$.

Hence

$$\text{breadth} = 16\text{ m}, \qquad \text{length} = 2(16) + 1 = 33\text{ m}.$$

Therefore, the dimensions of the prayer hall are $$33\text{ m} \times 16\text{ m}$$.

Step 1 — Identify the coefficients

The equation is already in standard form $$2x^2-4x+3=0$$, so

Step 2 — Write the discriminant formula

The discriminant of a quadratic equation $$ax^2+bx+c=0$$ is $$D=b^2-4ac$$.

Step 3 — Substitute and simplify

$$\begin{aligned}D&=(-4)^2-4(2)(3)\\ &=16-24\\ &=-8\end{aligned}$$

Step 4 — Interpret the discriminant

The value $$D=-8<0$$. When $$D<0$$, a quadratic equation has no real roots; instead, it possesses two unequal complex conjugate roots.

Let the two diametrically opposite gates be A and B and let the required point on the circular boundary be P.

Given the diameter $$AB = 13\,\text{m}$$, so the circular park has radius $$6.5\,\text{m}$$. Because A and B are the end-points of a diameter, the angle in a semicircle is a right angle; therefore

$$\angle APB = 90^\circ$$.

Denote $$PA = x\,\text{m}$$ and $$PB = y\,\text{m}$$. Two relations are obtained:

1. By the Pythagoras theorem in right △APB, $$x^2 + y^2 = AB^2 = 13^2 = 169$$.
2. The difference of the distances is given to be $$|PA - PB| = 7\,\text{m}$$. Without loss of generality assume $$PB > PA$$, so $$y - x = 7$$.

Substituting $$y = x + 7$$ from (2) into (1):

$$x^2 + (x + 7)^2 = 169$$
$$x^2 + x^2 + 14x + 49 = 169$$
$$2x^2 + 14x - 120 = 0$$
$$x^2 + 7x - 60 = 0$$.

The quadratic $$x^2 + 7x - 60 = 0$$ has discriminant $$D = 7^2 + 4 \times 60 = 49 + 240 = 289$$, so

$$x = \frac{-7 \pm \sqrt{289}}{2} = \frac{-7 \pm 17}{2}$$.

This yields $$x = 5\,\text{m}$$ or $$x = -12\,\text{m}$$. A distance cannot be negative, hence

$$PA = 5\,\text{m}$$.

Using $$y - x = 7$$, we have $$PB = 5 + 7 = 12\,\text{m}$$.

Conclusion. A pole can indeed be erected. There are two symmetrical positions of P (one on each side of diameter AB) where the distances to the gates are

$$PA = 5\,\text{m}, \; PB = 12\,\text{m}$$.

The given quadratic equation is

$$3x^2-2x+\dfrac13=0$$

Compare it with the standard form $$ax^2+bx+c=0$$.

• Coefficient of $$x^2$$ (that is, $$a$$): $$a = 3$$
• Coefficient of $$x$$ (that is, $$b$$): $$b = -2$$
• Constant term (that is, $$c$$): $$c = \dfrac13$$

Step 1 – Find the discriminant.

The discriminant of a quadratic equation is $$\Delta = b^2-4ac$$.

$$\begin{aligned} \Delta &= (-2)^2 - 4\,(3)\!\left(\dfrac13\right) \\ &= 4 - 4 \times 3 \times \dfrac13 \\ &= 4 - 4 \times 1 \\ &= 4 - 4 \\ &= 0 \end{aligned}$$

Step 2 – Decide the nature of the roots.

Because $$\Delta = 0$$, the quadratic equation has real and equal (coincident) roots.

Step 3 – Find the roots.

For $$\Delta = 0$$, both roots are given by

$$x = \frac{-b}{2a}$$

Substitute $$a = 3$$ and $$b = -2$$:

$$x = \frac{-(-2)}{2\times 3} = \frac{2}{6} = \frac13$$

Hence both roots are equal to $$\dfrac13$$.

Conclusion: The discriminant is $$0$$; therefore the equation has two equal real roots, each equal to $$\dfrac13$$.

The general form of a quadratic equation in one variable is $$ax^2 + bx + c = 0,$$ where $$a \neq 0$$. An equation is quadratic if, after all simplifications (expanding brackets, transposing terms, etc.), the highest power of the variable is 2 and the coefficient of that squared term is non-zero.

1. $$x^2 - 3x + 1 = 0$$ is already in the form $$ax^2 + bx + c = 0$$ with $$a = 1 \neq 0$$, $$b = -3$$, $$c = 1$$. Hence it is a quadratic equation.

2. $$3t(t - 3) + 1 = 0$$
   Expand the product:
   $$3t^2 - 9t + 1 = 0.$$
   This matches $$ax^2 + bx + c = 0$$ with $$a = 3 \neq 0$$. Therefore it is quadratic.

3. $$(x - 2)(x + 1) = x(x - 5)$$
   Expand both sides:
   Left: $$(x - 2)(x + 1) = x^2 - x - 2.$$
   Right: $$x(x - 5) = x^2 - 5x.$$
   Set the two expressions equal and simplify:
   $$x^2 - x - 2 = x^2 - 5x$$
   $$-x - 2 = -5x$$
   $$4x = 2 \;\;\Rightarrow\;\; x = \dfrac12.$$
   After cancellation the $$x^2$$ terms vanish, leaving a linear equation. Hence the original equation is not quadratic.

4. $$x^3 - 4x^2 + x = 0$$ already contains the term $$x^3$$ with non-zero coefficient. The highest power is 3. This is cubic, not quadratic.

5. $$(x - 1)^2 + 2 = 2(x - 1)$$
   Expand and simplify:
   $$(x^2 - 2x + 1) + 2 = 2x - 2$$
   $$x^2 - 2x + 3 = 2x - 2$$
   Bring all terms to one side:
   $$x^2 - 2x + 3 - 2x + 2 = 0$$
   $$x^2 - 4x + 5 = 0.$$
   Highest power is 2, and the coefficient of $$x^2$$ is 1 (non-zero). Thus it is a quadratic equation.

Summarising:

Let the two required numbers be $$x$$ and $$(27 - x)$$.

Their product is given to be $$182$$, so

$$x(27 - x) = 182$$

Expanding and arranging the terms:

$$27x - x^2 = 182$$

$$-x^2 + 27x - 182 = 0$$

Multiplying through by $$-1$$ to obtain a standard form:

$$x^2 - 27x + 182 = 0$$

We need two numbers whose product is $$182$$ and whose sum is $$27$$. Noting that $$13 \times 14 = 182$$ and $$13 + 14 = 27$$, we factorise:

$$x^2 - 27x + 182 = (x - 13)(x - 14) = 0$$

Thus,

$$x - 13 = 0 \;\; \text{or} \;\; x - 14 = 0$$

$$x = 13 \;\; \text{or} \;\; x = 14$$

If $$x = 13$$, the other number is $$27 - 13 = 14$$.
If $$x = 14$$, the other number is $$27 - 14 = 13$$.

Therefore, the two numbers are $$13$$ and $$14$$.

Let the smaller of the two consecutive positive integers be $$n$$.
Then the next integer is $$n+1$$.

According to the question,

$$n^{2} + (n+1)^{2} = 365$$

Expand and simplify:

$$n^{2} + n^{2} + 2n + 1 = 365$$

$$2n^{2} + 2n + 1 - 365 = 0$$

$$2n^{2} + 2n - 364 = 0$$

Divide by 2:

$$n^{2} + n - 182 = 0$$

To factorise, find two numbers whose product is $$-182$$ and sum is $$1$$: these are $$14$$ and $$-13$$.

$$n^{2} + 14n - 13n - 182 = 0$$

$$(n+14)(n-13) = 0$$

So, $$n+14 = 0$$ or $$n-13 = 0$$, giving $$n = -14$$ or $$n = 13$$.

Since the integers must be positive, $$n = 13$$.

Therefore the two consecutive positive integers are $$13$$ and $$14$$.

Let the base of the right-angled triangle be $$x\,\mathrm{cm}$$.

Then the altitude (height) is $$x-7\,\mathrm{cm}$$ (given it is 7 cm less than the base).

The hypotenuse is $$13\,\mathrm{cm}$$.

By the Pythagoras theorem,

$$x^{2}+(x-7)^{2}=13^{2}$$

Expanding and simplifying:

$$x^{2}+x^{2}-14x+49=169$$

$$2x^{2}-14x+49-169=0$$

$$2x^{2}-14x-120=0$$

Divide by $$2$$:

$$x^{2}-7x-60=0$$

Factorising the quadratic:

$$(x-12)(x+5)=0$$

Thus $$x=12$$ or $$x=-5$$.

Since a length cannot be negative, $$x=-5$$ is rejected.

Therefore the base is $$12\,\mathrm{cm}$$ and the altitude is $$x-7=5\,\mathrm{cm}$$.

Verification: $$12^{2}+5^{2}=169=13^{2}$$, confirming the result.

Let the number of pottery articles produced on the day be $$x$$.

The cost of making one article is “Rs. 3 more than twice the number of articles”, therefore

cost per article $$= 2x + 3$$ rupees.

Total cost of production:

$$x \times (2x + 3)$$

According to the question, this total cost equals Rs. 90:

$$x(2x + 3) = 90$$

Expanding gives the quadratic equation

$$2x^{2} + 3x - 90 = 0$$

To factorise, find two numbers whose product is $$2 \times (-90) = -180$$ and whose sum is $$3$$. The numbers $$15$$ and $$-12$$ satisfy these conditions.

Split the middle term:

$$2x^{2} + 15x - 12x - 90 = 0$$

Group and factor:

$$x(2x + 15) - 6(2x + 15) = 0$$

$$(2x + 15)(x - 6) = 0$$

Set each factor to zero:

$$2x + 15 = 0 \;\Longrightarrow\; x = -\dfrac{15}{2}\;(\text{rejected, negative})$$

$$x - 6 = 0 \;\Longrightarrow\; x = 6$$

Hence, the number of articles produced was $$6$$.

Cost per article:

$$2x + 3 = 2 \times 6 + 3 = 15$$ rupees.

Verification: $$6 \times 15 = 90$$ rupees, matching the given total cost.

Therefore, 6 pottery articles were produced and the cost of each article was Rs. 15.

Let the breadth of the rectangular mango grove be $$x\,\text{m}$$.

According to the statement, the length is twice the breadth, so

length $$=2x\,\text{m}$$.

Area of a rectangle $$=\text{length}\times\text{breadth}$$.

Thus, the given area condition is

$$2x\times x = 800$$

\(\Rightarrow\) $$2x^{2}=800$$

Divide by 2:

$$x^{2}=400$$

Rewrite as a quadratic equation:

$$x^{2}-400=0$$

Factorise:

$$(x-20)(x+20)=0$$

This gives two roots:

• $$x=20$$
• $$x=-20$$

A negative length has no physical meaning for a grove, so we reject $$x=-20$$.

Therefore,

• Breadth $$=x=20\,\text{m}$$
• Length $$=2x=40\,\text{m}$$

Check: $$\text{Area}=40\times20=800\,\text{m}^{2},$$ which matches the requirement.

Hence it is indeed possible to design such a mango grove.

Let the present ages of the two friends be $$x\text{ years}$$ and $$y\text{ years}$$.

According to the first condition,

$$x + y = 20 \quad\ldots(1)$$

Four years ago, their ages were $$x - 4$$ and $$y - 4$$. The second condition gives

$$(x - 4)(y - 4) = 48 \quad\ldots(2)$$

From (1), express $$y$$ in terms of $$x$$:

$$y = 20 - x$$

Substitute this in (2):

$$(x - 4)\bigl((20 - x) - 4\bigr) = 48$$

$$\Rightarrow (x - 4)(16 - x) = 48$$

Expand:

$$(x - 4)(16 - x) = x(16 - x) - 4(16 - x)$$

$$= 16x - x^2 - 64 + 4x$$

$$= -x^2 + 20x - 64$$

Set equal to $$48$$ as per the condition:

$$-x^2 + 20x - 64 = 48$$

Move all terms to one side:

$$-x^2 + 20x - 112 = 0$$

Multiply by $$-1$$ to make the quadratic standard:

$$x^2 - 20x + 112 = 0 \quad\ldots(3)$$

Find the discriminant $$\Delta$$ of (3):

$$\Delta = b^2 - 4ac = ( -20 )^2 - 4(1)(112) = 400 - 448 = -48$$

Because $$\Delta < 0$$, quadratic (3) has no real roots. Hence there is no real value of $$x$$ (and consequently of $$y$$) that satisfies both conditions simultaneously.

Therefore, the given situation is not possible.

Step 1: Assign variables
Let the length of the rectangular park be $$l ext{ m}$$ and its breadth be $$b ext{ m}$$.

Step 2: Translate the perimeter condition
The perimeter of a rectangle is given by $$2(l+b)$$. We are told it must be $$80 ext{ m}$$, so
$$2(l+b)=80 \\[2pt]\ \Rightarrow l+b=40 \\[2pt]\ \Rightarrow b=40-l.$$

Step 3: Translate the area condition
The area is given by $$l\,b$$ and must be $$400 \text{ m}^2$$:
$$l\,b = 400.$$
Substitute $$b=40-l$$ from Step 2:
$$l(40-l)=400.$$

Step 4: Form and simplify the quadratic equation
Expand and bring all terms to one side:
$$40l-l^2=400 \\[2pt]\ \Rightarrow -l^2+40l-400=0 \\[2pt]\ \Rightarrow l^2-40l+400=0.$$

Step 5: Solve the quadratic
The quadratic equation is $$l^2-40l+400=0$$ with
$$a=1, \; b=-40, \; c=400.$$
Its discriminant is
$$\Delta=b^2-4ac=(-40)^2-4(1)(400)=1600-1600=0.$$
Because $$\Delta=0$$, the quadratic has one repeated (real) root:
$$l=\frac{-b}{2a}=\frac{-(-40)}{2\times1}=20.$$

Step 6: Find the breadth
Using $$b=40-l$$:
$$b=40-20=20.$$

Step 7: Interpret the result
The positive solution $$l=20,\; b=20$$ satisfies both the perimeter and area requirements. Hence, a square park of side $$20 \text{ m}$$ can indeed be designed.

Verification (optional)
Perimeter: $$2(20+20)=80 \text{ m}\;\;\checkmark$$
Area: $$20\times20=400 \text{ m}^2\;\;\checkmark$$

Therefore, it is possible, and the park must be a square measuring 20 m × 20 m.