Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We would like to find out the number of toys produced on that day.
(i) Let the number of marbles John had be x. Then the number of marbles Jivanti had is \(45-x\) (since together they have 45).
When each of them loses 5 marbles: John is left with \(x-5\), and Jivanti is left with \(45-x-5=40-x\).
Their product is given as 124:
Expanding the left side:
Rearranging into the standard form \(ax^{2}+bx+c=0\) (multiplying throughout by \(-1\)):
This is the required quadratic equation. (It is solved in Exercise 4.2, Q2(i) below.)
(ii) Let the number of toys produced on that day be x. Then the cost of production of each toy (in rupees) is \(55-x\).
Total cost = (number of toys) × (cost per toy) = ₹750:
Rearranging into standard form (multiplying throughout by \(-1\)):
This is the required quadratic equation. (It is solved in Exercise 4.2, Q2(ii) below.)
Check whether the following are quadratic equations:
(i) \((x-2)^{2}+1=2x-3\) (ii) \(x(x+1)+8=(x+2)(x-2)\)
(iii) \(x(2x+3)=x^{2}+1\) (iv) \((x+2)^{3}=x^{3}-4\)
A given equation is a quadratic equation only if, once every term is expanded and like terms are collected, the \(x^{2}\) term survives with a non-zero coefficient. So each part must be expanded fully before deciding.
(i) LHS \(=(x-2)^{2}+1=x^{2}-4x+4+1=x^{2}-4x+5\). So the equation becomes:
This is of the form \(ax^{2}+bx+c=0\) with \(a=1\ne0\). It is a quadratic equation.
(ii) LHS \(=x(x+1)+8=x^{2}+x+8\). RHS \(=(x+2)(x-2)=x^{2}-4\). So the equation becomes:
The \(x^{2}\) terms cancel from both sides, leaving:
This is linear, not of the form \(ax^{2}+bx+c=0\) with \(a\ne0\). It is NOT a quadratic equation (even though it looked like one before expanding).
(iii) LHS \(=x(2x+3)=2x^{2}+3x\). So \(x(2x+3)=x^{2}+1\) becomes:
This is of the form \(ax^{2}+bx+c=0\) with \(a=1\ne0\). It is a quadratic equation.
(iv) LHS \(=(x+2)^{3}=x^{3}+3x^{2}(2)+3x(2)^{2}+2^{3}=x^{3}+6x^{2}+12x+8\). So \((x+2)^{3}=x^{3}-4\) becomes:
The \(x^{3}\) terms cancel from both sides, leaving:
This is of the form \(ax^{2}+bx+c=0\) with \(a=1\ne0\). It is a quadratic equation — even though it looked like a cubic equation (degree 3) before expanding.
The lesson of this example: always simplify fully before deciding whether an equation is quadratic — simplification can turn an apparent quadratic into a linear equation (part ii), or an apparent cubic into a genuine quadratic (part iv).
Find the roots of the equation \(2x^{2}-5x+3=0\), by factorisation.
To factorise \(2x^{2}-5x+3\) by splitting the middle term, we need two numbers whose sum is \(-5\) (the coefficient of x) and whose product is \(2\times3=6\) (the coefficient of \(x^{2}\) times the constant term).
Those two numbers are \(-2\) and \(-3\), since \((-2)+(-3)=-5\) and \((-2)\times(-3)=6\). So split the middle term \(-5x\) as \(-2x-3x\):
Now factor by grouping:
So the equation \(2x^{2}-5x+3=0\) becomes \((2x-3)(x-1)=0\), which is true exactly when one of the two factors is zero:
So the roots of \(2x^{2}-5x+3=0\) are \(x=\dfrac{3}{2}\) and \(x=1\).
Verification: at \(x=1\): \(2(1)^{2}-5(1)+3=2-5+3=0\ \checkmark\). At \(x=\tfrac{3}{2}\): \(2\left(\tfrac{3}{2}\right)^{2}-5\left(\tfrac{3}{2}\right)+3=\tfrac{9}{2}-\tfrac{15}{2}+3=-3+3=0\ \checkmark\).
Find the roots of the quadratic equation \(6x^{2}-x-2=0\).
We need two numbers whose sum is \(-1\) (the coefficient of x) and whose product is \(6\times(-2)=-12\) (coefficient of \(x^{2}\) times the constant term). Those numbers are \(3\) and \(-4\), since \(3+(-4)=-1\) and \(3\times(-4)=-12\). Split \(-x\) as \(3x-4x\):
Factor by grouping:
So \(6x^{2}-x-2=0\) becomes \((3x-2)(2x+1)=0\):
So the roots of \(6x^{2}-x-2=0\) are \(\dfrac{2}{3}\) and \(-\dfrac{1}{2}\).
Verification: at \(x=\tfrac{2}{3}\): \(6\left(\tfrac{4}{9}\right)-\tfrac{2}{3}-2=\tfrac{8}{3}-\tfrac{2}{3}-2=2-2=0\ \checkmark\). At \(x=-\tfrac{1}{2}\): \(6\left(\tfrac{1}{4}\right)-\left(-\tfrac{1}{2}\right)-2=\tfrac{3}{2}+\tfrac{1}{2}-2=2-2=0\ \checkmark\).
Find the roots of the quadratic equation \(3x^{2}-2\sqrt{6}\,x+2=0\).
Here the middle term must be split using surds. We need two terms whose sum is \(-2\sqrt{6}\,x\) and whose product is \(3x^{2}\times2=6x^{2}\). Since \(\sqrt{6}=\sqrt{2}\times\sqrt{3}\), try \(-\sqrt{6}x\) and \(-\sqrt{6}x\) (their sum is \(-2\sqrt{6}x\) and their product is \(6x^{2}\)):
Writing \(3x^{2}=\sqrt{3}\,x\times\sqrt{3}\,x\) and \(\sqrt{6}=\sqrt{2}\times\sqrt{3}\), factor by grouping:
So the equation becomes \(\left(\sqrt{3}\,x-\sqrt{2}\right)^{2}=0\), which gives the same equation twice:
Since both factors are identical, this root is repeated — one root for each of the two (equal) factors. So the roots of \(3x^{2}-2\sqrt{6}\,x+2=0\) are:
Verification: \(\sqrt{\tfrac{2}{3}}=\tfrac{\sqrt{6}}{3}\). Substituting: \(3\left(\tfrac{6}{9}\right)-2\sqrt{6}\left(\tfrac{\sqrt{6}}{3}\right)+2=2-4+2=0\ \checkmark\).
Find the dimensions of the prayer hall discussed in Section 4.1 (a charity trust wants to build a hall with a carpet area of 300 square metres, whose length is one metre more than twice its breadth).
In Section 4.1, taking the breadth of the hall to be x m, its length works out to \((2x+1)\) m, and the condition “area \(=300\ \text{m}^{2}\)” gives the quadratic equation \(2x^{2}+x-300=0\). We now solve it by factorisation.
We need two numbers whose sum is \(1\) (the coefficient of x) and whose product is \(2\times(-300)=-600\). Those numbers are \(25\) and \(-24\), since \(25+(-24)=1\) and \(25\times(-24)=-600\). Split \(x\) as \(25x-24x\):
Factor by grouping:
So \(2x^{2}+x-300=0\) becomes \((x-12)(2x+25)=0\):
Since x is the breadth of the hall, it cannot be negative — so \(x=-12.5\) is rejected. Thus the breadth is 12 m, and the length is \(2(12)+1=\)25 m.
Find the discriminant of the quadratic equation \(2x^{2}-4x+3=0\), and hence find the nature of its roots.
The given equation is of the form \(ax^{2}+bx+c=0\), with \(a=2,\ b=-4,\ c=3\). The discriminant is:
Since the discriminant is negative (\(-8<0\)), the given equation has no real roots.
A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Let P be the required location of the pole, and let the distance of the pole from gate B be x m, i.e., \(BP=x\) m. Since the difference of the distances from the two gates is 7 m and \(AP>BP\), \(AP-BP=7\), so \(AP=(x+7)\) m.
Now \(AB=13\) m, and since AB is a diameter of the circle, the angle in the semicircle at P is a right angle: \(\angle APB=90^{\circ}\). By the Pythagoras theorem:
Dividing throughout by 2, the distance x satisfies:
It would be possible to place the pole only if this equation has real roots. Its discriminant is:
Since the discriminant is positive, the equation has two real roots, so it IS possible to erect the pole. Solving by the quadratic formula:
Since x is a distance, it must be positive, so \(x=-12\) is rejected. Thus \(x=5\).
The pole has to be erected on the boundary of the park at a distance of 5 m from gate B and \(x+7=12\) m, i.e. 12 m from gate A.
Verification: \(5^{2}+12^{2}=25+144=169=13^{2}\ \checkmark\), and \(12-5=7\ \checkmark\).
Find the discriminant of the equation \(3x^{2}-2x+\dfrac{1}{3}=0\) and hence find the nature of its roots. Find them, if they are real.
Here \(a=3,\ b=-2,\ c=\dfrac{1}{3}\). The discriminant is:
Since the discriminant is zero, the given quadratic equation has two equal real roots.
When the discriminant is 0, both roots equal \(\dfrac{-b}{2a}\):
So the roots are \(\dfrac{1}{3}\) and \(\dfrac{1}{3}\).
Verification: \(3\left(\tfrac{1}{3}\right)^{2}-2\left(\tfrac{1}{3}\right)+\tfrac{1}{3}=\tfrac{1}{3}-\tfrac{2}{3}+\tfrac{1}{3}=0\ \checkmark\).
Check whether the following are quadratic equations:
(i) \((x+1)^{2}=2(x-3)\) (ii) \(x^{2}-2x=(-2)(3-x)\)
(iii) \((x-2)(x+1)=(x-1)(x+3)\) (iv) \((x-3)(2x+1)=x(x+5)\)
(v) \((2x-1)(x-3)=(x+5)(x-1)\) (vi) \(x^{2}+3x+1=(x-2)^{2}\)
(vii) \((x+2)^{3}=2x(x^{2}-1)\) (viii) \(x^{3}-4x^{2}-x+1=(x-2)^{3}\)
(i) \((x+1)^{2}=2(x-3)\ \Rightarrow\ x^{2}+2x+1=2x-6\ \Rightarrow\ x^{2}+7=0\). This is of the form \(ax^{2}+bx+c=0\) (with \(b=0\), which is still allowed since \(a\ne0\)). Quadratic.
(ii) \(x^{2}-2x=(-2)(3-x)=-6+2x\ \Rightarrow\ x^{2}-4x+6=0\). Quadratic.
(iii) LHS \(=(x-2)(x+1)=x^{2}-x-2\). RHS \(=(x-1)(x+3)=x^{2}+2x-3\). Equating: \(x^{2}-x-2=x^{2}+2x-3\ \Rightarrow\ 3x-1=0\). The \(x^{2}\) terms cancel — this is linear. NOT quadratic.
(iv) LHS \(=(x-3)(2x+1)=2x^{2}-5x-3\). RHS \(=x(x+5)=x^{2}+5x\). Equating: \(2x^{2}-5x-3=x^{2}+5x\ \Rightarrow\ x^{2}-10x-3=0\). Quadratic.
(v) LHS \(=(2x-1)(x-3)=2x^{2}-7x+3\). RHS \(=(x+5)(x-1)=x^{2}+4x-5\). Equating: \(2x^{2}-7x+3=x^{2}+4x-5\ \Rightarrow\ x^{2}-11x+8=0\). Quadratic.
(vi) RHS \(=(x-2)^{2}=x^{2}-4x+4\). Equating: \(x^{2}+3x+1=x^{2}-4x+4\ \Rightarrow\ 7x-3=0\). The \(x^{2}\) terms cancel — linear. NOT quadratic.
(vii) LHS \(=(x+2)^{3}=x^{3}+6x^{2}+12x+8\). RHS \(=2x(x^{2}-1)=2x^{3}-2x\). Equating: \(x^{3}+6x^{2}+12x+8=2x^{3}-2x\ \Rightarrow\ x^{3}-6x^{2}-14x-8=0\). The \(x^{3}\) terms do NOT cancel (coefficients 1 and 2 differ) — this is a genuine cubic. NOT quadratic.
(viii) RHS \(=(x-2)^{3}=x^{3}-6x^{2}+12x-8\). Equating: \(x^{3}-4x^{2}-x+1=x^{3}-6x^{2}+12x-8\). The \(x^{3}\) terms cancel (both sides have coefficient 1), leaving:
This IS of the form \(ax^{2}+bx+c=0\) with \(a=2\ne0\), even though it looked cubic at first glance. Quadratic.
Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
(i) Let the breadth of the plot be x m. Then the length is \((2x+1)\) m (one more than twice the breadth). Area \(=\) length \(\times\) breadth \(=528\):
This is the required quadratic equation, in the breadth x.
(ii) Let the smaller of the two consecutive positive integers be x. Then the next one is \((x+1)\). Their product is 306:
This is the required quadratic equation.
(iii) Let Rohan’s present age be x years. Then his mother’s present age is \((x+26)\) years. Three years from now, Rohan’s age will be \((x+3)\) and his mother’s will be \((x+26+3)=(x+29)\). Their product will be 360:
This is the required quadratic equation, in Rohan’s present age x.
(iv) Let the uniform speed of the train be x km/h (\(x>8\)). Time taken to cover 480 km at this speed \(=\dfrac{480}{x}\) hours. At the reduced speed \((x-8)\) km/h, the time taken would be \(\dfrac{480}{x-8}\) hours, which is 3 hours MORE:
Multiplying throughout by the common denominator \(x(x-8)\):
Dividing throughout by 3:
This is the required quadratic equation, in the speed x.
Find the roots of the following quadratic equations by factorisation:
(i) \(x^{2}-3x-10=0\) (ii) \(2x^{2}+x-6=0\)
(iii) \(\sqrt{2}\,x^{2}+7x+5\sqrt{2}=0\) (iv) \(2x^{2}-x+\dfrac{1}{8}=0\)
(v) \(100x^{2}-20x+1=0\)
(i) Need two numbers with sum \(-3\) and product \(-10\): those are \(-5\) and \(2\). Split \(-3x=-5x+2x\):
So \(x=5\) or \(x=-2\). Roots: 5 and \(-2\).
(ii) Need two numbers with sum \(1\) and product \(2\times(-6)=-12\): those are \(4\) and \(-3\). Split \(x=4x-3x\):
So \(x=-2\) or \(x=\dfrac{3}{2}\). Roots: \(-2\) and \(\dfrac{3}{2}\).
(iii) Need two terms with sum \(7x\) and product \(\sqrt{2}x^{2}\times5\sqrt{2}=10x^{2}\): those are \(2x\) and \(5x\). Split \(7x=2x+5x\):
Writing \(\sqrt{2}x^{2}=\sqrt{2}x\times x\) and \(2x=\sqrt{2}x\times\sqrt{2}\), factor the first pair by \(\sqrt{2}x\) and the second pair by \(5\):
So \(x=-\sqrt{2}\) or \(\sqrt{2}x=-5\Rightarrow x=-\dfrac{5}{\sqrt{2}}=-\dfrac{5\sqrt{2}}{2}\). Roots: \(-\sqrt{2}\) and \(-\dfrac{5\sqrt{2}}{2}\).
(iv) Multiply throughout by 8 to clear the fraction (a valid step, since multiplying both sides of an equation by a non-zero constant does not change its roots):
Need two numbers with sum \(-8\) and product \(16\times1=16\): those are \(-4\) and \(-4\). Split \(-8x=-4x-4x\):
Both factors are identical, so the root is repeated: \(4x=1\Rightarrow x=\dfrac{1}{4}\). Roots: \(\dfrac{1}{4}\) and \(\dfrac{1}{4}\).
(v) Need two numbers with sum \(-20\) and product \(100\times1=100\): those are \(-10\) and \(-10\). Split \(-20x=-10x-10x\):
So \(10x=1\Rightarrow x=\dfrac{1}{10}\), repeated. Roots: \(\dfrac{1}{10}\) and \(\dfrac{1}{10}\).
Solve the problems given in Example 1 (in the Worked Examples above).
(i) John and Jivanti’s marbles: Example 1(i) found the equation \(x^{2}-45x+324=0\), where x is the number of marbles John had. Need two numbers with sum \(-45\) and product \(324\): those are \(-36\) and \(-9\). Split \(-45x=-36x-9x\):
So \(x=36\) or \(x=9\). Both are valid (x must be a positive integer with \(5\le x\le40\), so that neither “\(x-5\)” nor “\(40-x\)” is negative — both 36 and 9 satisfy this). So John had 36 marbles and Jivanti had 9, OR John had 9 marbles and Jivanti had 36 — the two roots simply correspond to which of the two friends is “John” in the story.
(ii) The cottage industry’s toys: Example 1(ii) found the equation \(x^{2}-55x+750=0\), where x is the number of toys produced. Need two numbers with sum \(-55\) and product \(750\): since \(750=25\times30\) and \(25+30=55\), those are \(-25\) and \(-30\). Split \(-55x=-25x-30x\):
So \(x=25\) or \(x=30\). Both give a positive cost per toy (\(55-25=30\) or \(55-30=25\), both positive), so both are valid: 25 toys produced (at ₹30 each), OR 30 toys produced (at ₹25 each) — either reading is consistent with a total cost of \(25\times30=750\).
Find two numbers whose sum is 27 and product is 182.
Let one number be x. Since the two numbers add up to 27, the other is \((27-x)\). Their product is 182:
Need two numbers with sum \(-27\) and product \(182\): since \(182=13\times14\) and \(13+14=27\), those are \(-13\) and \(-14\). Split \(-27x=-13x-14x\):
So \(x=13\) or \(x=14\). The other number is \(27-x\), i.e. 14 or 13 respectively. So the two numbers are 13 and 14.
Verification: \(13+14=27\ \checkmark\) and \(13\times14=182\ \checkmark\).
Find two consecutive positive integers, sum of whose squares is 365.
Let the smaller positive integer be x, so the next consecutive integer is \((x+1)\). The sum of their squares is 365:
Dividing throughout by 2:
Need two numbers with sum \(1\) and product \(-182\): those are \(14\) and \(-13\). Split \(x=14x-13x\):
So \(x=-14\) or \(x=13\). Since x must be a positive integer, \(x=-14\) is rejected. So \(x=13\), and the two consecutive integers are 13 and 14.
Verification: \(13^{2}+14^{2}=169+196=365\ \checkmark\).
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let the base be x cm. Then the altitude is \((x-7)\) cm. By the Pythagoras theorem, since the hypotenuse is 13 cm:
Dividing throughout by 2:
Need two numbers with sum \(-7\) and product \(-60\): those are \(-12\) and \(5\). Split \(-7x=-12x+5x\):
So \(x=12\) or \(x=-5\). Since x is the base of a triangle, it must be positive, so \(x=-5\) is rejected. So \(x=12\), and the altitude is \(12-7=5\).
The other two sides are base \(=12\) cm and altitude \(=5\) cm.
Verification: \(5^{2}+12^{2}=25+144=169=13^{2}\ \checkmark\).
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Let the number of articles produced be x. Then the cost of production of each article (in rupees) is \((2x+3)\) — 3 more than twice the number of articles. Total cost \(=\) (number of articles) \(\times\) (cost per article) \(=90\):
Need two numbers with sum \(3\) and product \(2\times(-90)=-180\): those are \(15\) and \(-12\). Split \(3x=15x-12x\):
So \(x=-\dfrac{15}{2}\) or \(x=6\). Since x is a number of articles, it must be a positive integer, so \(x=-\dfrac{15}{2}\) is rejected. So \(x=6\).
The number of articles produced is 6, and the cost of each article is \(2(6)+3=\)₹15.
Verification: \(6\times15=90\ \checkmark\).
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) \(2x^{2}-3x+5=0\) (ii) \(3x^{2}-4\sqrt{3}\,x+4=0\)
(iii) \(2x^{2}-6x+3=0\)
(i) Here \(a=2,\ b=-3,\ c=5\). Discriminant:
Since the discriminant is negative (\(-31<0\)), the equation has no real roots.
(ii) Here \(a=3,\ b=-4\sqrt{3},\ c=4\). Discriminant:
Since the discriminant is zero, the equation has two equal real roots, both equal to \(\dfrac{-b}{2a}\):
Roots: \(\dfrac{2\sqrt{3}}{3}\) and \(\dfrac{2\sqrt{3}}{3}\).
(iii) Here \(a=2,\ b=-6,\ c=3\). Discriminant:
Since the discriminant is positive (\(12>0\)), the equation has two distinct real roots, given by the quadratic formula:
Roots: \(\dfrac{3+\sqrt{3}}{2}\) and \(\dfrac{3-\sqrt{3}}{2}\).
Find the values of k for each of the following quadratic equations, so that they have two equal roots:
(i) \(2x^{2}+kx+3=0\) (ii) \(kx(x-2)+6=0\)
(i) Here \(a=2,\ b=k,\ c=3\). For two equal roots, the discriminant must be zero:
So \(k=2\sqrt{6}\) or \(k=-2\sqrt{6}\).
(ii) First expand into standard form: \(kx(x-2)+6=0\ \Rightarrow\ kx^{2}-2kx+6=0\), so \(a=k,\ b=-2k,\ c=6\). For two equal roots, the discriminant must be zero:
So \(k=0\) or \(k=6\). But \(k=0\) must be rejected: if \(k=0\), the coefficient of \(x^{2}\) becomes 0, the equation collapses to \(6=0\) (never true, and not even an equation in x any more) — it stops being a quadratic equation at all, which is exactly the condition \(a\ne0\) in \(ax^{2}+bx+c=0\) that we started from. So the only valid answer is \(k=6\).
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Let the breadth be x m. Then the length is \(2x\) m (twice the breadth). Area \(=\) length \(\times\) breadth \(=800\):
Here \(a=1,\ b=0,\ c=-400\). Discriminant:
Since the discriminant is positive, the equation has two real roots — so yes, it is possible to design such a grove. Solving:
Since x is the breadth, it must be positive, so \(x=-20\) is rejected. Breadth \(=20\) m, length \(=2(20)=40\) m.
So it is possible, with breadth 20 m and length 40 m.
Verification: \(40\times20=800\ \checkmark\).
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages (in years) was 48.
Let the present age of one friend be x years. Since their ages sum to 20, the other friend’s present age is \((20-x)\). Four years ago, their ages were \((x-4)\) and \((20-x-4)=(16-x)\), and the product of those was 48:
Rearranging into standard form (multiplying by \(-1\)):
Here \(a=1,\ b=-20,\ c=112\). Discriminant:
Since the discriminant is negative (\(-48<0\)), this equation has no real roots at all. So the situation described is NOT possible — there is no real age x for which this could happen; it is impossible for two friends with these age and (four-years-ago) product conditions to exist.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.
Let the length be x m. Since the perimeter is 80 m, length \(+\) breadth \(=40\) m, so the breadth is \((40-x)\) m. Area \(=\) length \(\times\) breadth \(=400\):
Rearranging into standard form:
Here \(a=1,\ b=-40,\ c=400\). Discriminant:
Since the discriminant is zero, the equation has two equal real roots — so yes, it is possible. Both roots equal \(\dfrac{-b}{2a}\):
So length \(=20\) m, and breadth \(=40-20=20\) m as well — the park is a square of side 20 m.
Verification: perimeter \(=2(20+20)=80\ \checkmark\); area \(=20\times20=400\ \checkmark\).
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