Quadratic Equations

Check whether the following is quadratic equation :

(x + 1)² = 2(x - 3)

Answer

Given,

Equation : (x + 1)² = 2(x - 3)

Solving above equation,

⇒ x² + 1 + 2x = 2x - 6

⇒ x² = 2x - 6 - 2x - 1

⇒ x² = -7

⇒ x² + 7 = 0.

The above equation is of the form ax² + bx + c = 0.

Hence, the above equation is a quadratic equation.

Check whether the following is quadratic equation :

x² - 2x = (-2)(3 - x)

Answer

Given,

Equation : x² - 2x = (-2)(3 - x)

Solving above equation,

⇒ x² - 2x = -6 + 2x

⇒ x² - 2x - 2x + 6 = 0

⇒ x² - 4x + 6 = 0.

The above equation is of the form ax² + bx + c = 0.

Hence, the above equation is a quadratic equation.

Check whether the following is quadratic equation :

(x - 2)(x + 1) = (x - 1)(x + 3)

Answer

Given,

Equation : (x - 2)(x + 1) = (x - 1)(x + 3)

⇒ x² + x - 2x - 2 = x² + 3x - x - 3

⇒ x² - x - 2 = x² + 2x - 3

⇒ x² - x² = 2x + x - 3 + 2

⇒ 3x - 1 = 0.

The above equation is not of the form ax² + bx + c = 0.

Hence, the above equation is not a quadratic equation.

Check whether the following is quadratic equation :

(x - 3)(2x + 1) = x(x + 5)

Answer

Given,

Equation : (x - 3)(2x + 1) = x(x + 5)

⇒ 2x² + x - 6x - 3 = x² + 5x

⇒ 2x² - x² + x - 6x - 5x - 3 = 0

⇒ x² - 10x - 3 = 0.

The above equation is of the form ax² + bx + c = 0.

Hence, the above equation is a quadratic equation.

Check whether the following is quadratic equation :

(2x - 1)(x - 3) = (x + 5)(x - 1)

Answer

Given,

Equation : (2x - 1)(x - 3) = (x + 5)(x - 1)

⇒ 2x² - 6x - x + 3 = x² - x + 5x - 5

⇒ 2x² - 7x + 3 = x² + 4x - 5

⇒ 2x² - x² - 7x - 4x + 3 + 5 = 0

⇒ x² - 11x + 8 = 0.

The above equation is of the form ax² + bx + c = 0.

Hence, the above equation is a quadratic equation.

Check whether the following is quadratic equation :

x² + 3x + 1 = (x - 2)²

Answer

Given,

Equation : x² + 3x + 1 = (x - 2)²

⇒ x² + 3x + 1 = x² + 4 - 4x

⇒ x² - x² + 3x + 4x + 1 - 4 = 0

⇒ 7x - 3 = 0.

The above equation is not of the form ax² + bx + c = 0.

Hence, the above equation is not a quadratic equation.

Check whether the following is quadratic equation :

(x + 2)³ = 2x(x² - 1)

Answer

Given,

Equation : (x + 2)³ = 2x(x² - 1)

⇒ x³ + 2³ + 3 × x × 2(x + 2) = 2x³ - 2x

⇒ x³ + 8 + 6x(x + 2) = 2x³ - 2x

⇒ x³ + 8 + 6x² + 12x = 2x³ - 2x

⇒ 2x³ - x³ - 6x² - 2x - 8 = 0

⇒ x³ - 6x² - 2x - 8 = 0.

The above equation is not of the form ax² + bx + c = 0.

Hence, the above equation is not a quadratic equation.

Check whether the following is quadratic equation :

x³ - 4x² - x + 1 = (x - 2)³

Answer

Given,

Equation : x³ - 4x² - x + 1 = (x - 2)³

⇒ x³ - 4x² - x + 1 = x³ - 2³ - 3.x.2(x - 2)

⇒ x³ - 4x² - x + 1 = x³ - 8 - 6x(x - 2)

⇒ x³ - x³ - 4x² - x + 1 + 8 = -6x² + 12x

⇒ -4x² + 6x² - x - 12x + 9 = 0

⇒ 2x² - 13x + 9 = 0.

The above equation is of the form ax² + bx + c = 0.

Hence, the above equation is a quadratic equation.

Represent the following situation in the form of quadratic equation :

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.

Answer

Let breadth of rectangular plot be x meters.

According to question,

Length = (2x + 1) meters

By formula,

Area = Length × Breadth

Substituting values we get :

⇒ 528 = x(2x + 1)

⇒ 2x² + x = 528

⇒ 2x² + x - 528 = 0

Hence, required quadratic equation is 2x² + x - 528 = 0.

Represent the following situation in the form of quadratic equation :

The product of two consecutive positive integers is 306. We need to find the integers.

Answer

Let first number be x and second be (x + 1).

Given,

Product of two consecutive positive integers is 306.

∴ x(x + 1) = 306

⇒ x² + x = 306

⇒ x² + x - 306 = 0.

Hence, required quadratic equation is x² + x - 306 = 0.

Represent the following situation in the form of quadratic equation :

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.

Answer

Let Rohan's present age be x years.

So, Mother's age = (x + 26) years.

Age of Rohan after 3 years = (x + 3) years.

Age of Rohan's mother after 3 years = (x + 26 + 3) = (x + 29) years.

Given,

Product of their ages after 3 years = 360.

⇒ (x + 3)(x + 29) = 360

⇒ x² + 29x + 3x + 87 = 360

⇒ x² + 32x + 87 - 360 = 0

⇒ x² + 32x - 273 = 0.

Hence, required quadratic equation is x² + 32x - 273 = 0.

Represent the following situation in the form of quadratic equation :

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.

Answer

Let speed of train be x km/hour.

By formula,

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance.

$\therefore \dfrac{480}{x - 8} - \dfrac{480}{x} = 3 \\[1em] \Rightarrow \dfrac{480x - 480(x - 8)}{x(x - 8)} = 3 \\[1em] \Rightarrow \dfrac{480x - 480x + 3840}{x^2 - 8x} = 3 \\[1em] \Rightarrow 3840 = 3(x^2 - 8x) \\[1em] \Rightarrow 3x^2 - 24x = 3840 \\[1em] \Rightarrow 3(x^2 - 8x) = 3840 \\[1em] \Rightarrow x^2 - 8x = \dfrac{3840}{3} \\[1em] \Rightarrow x^2 - 8x = 1280 \\[1em] \Rightarrow x^2 - 8x - 1280 = 0.$

Hence, the required equation is x² - 8x - 1280 = 0.

Find the roots of the following quadratic equation by factorisation :

x² - 3x - 10 = 0

Answer

Solving,

⇒ x² - 3x - 10 = 0

⇒ x² - 5x + 2x - 10 = 0

⇒ x(x - 5) + 2(x - 5) = 0

⇒ (x + 2)(x - 5) = 0

⇒ x + 2 = 0 or x - 5 = 0

⇒ x = -2 or x = 5.

Hence, x = -2 or x = 5.

Find the roots of the following quadratic equation by factorisation :

2x² + x - 6 = 0

Answer

Solving,

⇒ 2x² + x - 6 = 0

⇒ 2x² + 4x - 3x - 6 = 0

⇒ 2x(x + 2) - 3(x + 2) = 0

⇒ (2x - 3)(x + 2) = 0

⇒ 2x - 3 = 0 or x + 2 = 0

⇒ 2x = 3 or x = -2

⇒ x = $\dfrac{3}{2}$ or x = -2.

Hence, x = -2 or x = $\dfrac{3}{2}$.

Find the roots of the following quadratic equation by factorisation :

$\sqrt{2}x^2 +7x + 5\sqrt{2} = 0$

Answer

Solving,

$\Rightarrow \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \\[1em] \Rightarrow \sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0 \\[1em] \Rightarrow \sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \\[1em] \Rightarrow (\sqrt{2}x + 5)(x + \sqrt{2}) = 0 \\[1em] \Rightarrow \sqrt{2}x + 5 = 0 \text{ or } x + \sqrt{2} = 0 \\[1em] \Rightarrow \sqrt{2}x = -5 \text{ or } x = -\sqrt{2} \\[1em] \Rightarrow x = -\dfrac{5}{\sqrt{2}} \text{ or } x = -\sqrt{2}.$

Hence, x = $-\dfrac{5}{\sqrt{2}}\text{ or x} = -\sqrt{2}$.

Find the roots of the following quadratic equations by factorisation :

$2x^2 - x + \dfrac{1}{8} = 0$

Answer

Solving,

$\Rightarrow 2x^2 - x + \dfrac{1}{8} = 0 \\[1em] \Rightarrow \dfrac{16x^2 - 8x + 1}{8} = 0 \\[1em] \Rightarrow 16x^2 - 8x + 1 = 0 \\[1em] \Rightarrow 16x^2 - 4x - 4x + 1 = 0 \\[1em] \Rightarrow 4x(4x - 1) - 1(4x - 1) = 0 \\[1em] \Rightarrow (4x - 1)(4x - 1) = 0 \\[1em] \Rightarrow 4x - 1 = 0 \text{ or } 4x - 1 = 0 \\[1em] \Rightarrow 4x = 1 \text{ or } 4x = 1 \\[1em] \Rightarrow x = \dfrac{1}{4} \text{ or } x =\dfrac{1}{4}.$

Hence, x = $\dfrac{1}{4}, \dfrac{1}{4}$.

Find the roots of the following quadratic equations by factorisation :

100x² - 20x + 1 = 0

Answer

Solving,

⇒ 100x² - 20x + 1 = 0

⇒ 100x² - 10x - 10x + 1 = 0

⇒ 10x(10x - 1) - 1(10x - 1) = 0

⇒ (10x - 1)(10x - 1) = 0

⇒ 10x - 1 = 0 or 10x - 1 = 0

⇒ 10x = 1 or 10x = 1

⇒ x = $\dfrac{1}{10} \text{ or x} = \dfrac{1}{10}$.

Hence, x = $\dfrac{1}{10}, \dfrac{1}{10}$.

Solve :

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.

Answer

Let initially John had x marbles, so Jivanti had (45 - x) marbles.

After loosing 5 marbles by each of them, they have :

John = (x - 5), Jivanti = (45 - x - 5) = (40 - x).

Given,

Product of marbles they now have is 124.

∴ (x - 5)(40 - x) = 124

⇒ 40x - x² - 200 + 5x = 124

⇒ 45x - x² - 200 - 124 = 0

⇒ 45x - x² - 324 = 0

⇒ x² - 45x + 324 = 0

⇒ x² - 9x - 36x + 324 = 0

⇒ x(x - 9) - 36(x - 9) = 0

⇒ (x - 36)(x - 9) = 0

⇒ x - 36 = 0 or x - 9 = 0

⇒ x = 36 or x = 9.

Hence, initially they start with 9 and 36 marbles.

Solve :

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.

Answer

Let no. of toys produced be x.

Cost of each toy = ₹ (55 - x)

Given,

Total cost = 750

⇒ x(55 - x) = 750

⇒ 55x - x² = 750

⇒ x² - 55x + 750 = 0

⇒ x² - 30x - 25x + 750 = 0

⇒ x(x - 30) - 25(x - 30) = 0

⇒ (x - 25)(x - 30) = 0

⇒ x - 25 = 0 or x - 30 = 0

⇒ x = 25 or x = 30.

If x = 25, 55 - x = 30 and,

if x = 30, 55 - x = 35.

Hence, if no. of articles produced are 25 then cost of each article = ₹ 30 and if no. of articles produced are 30 then cost of each article = ₹ 25.

Find two numbers whose sum is 27 and product is 182.

Answer

Let first number be x and second number = (27 - x).

Given,

Product = 182

∴ x(27 - x) = 182

⇒ 27x - x² = 182

⇒ x² - 27x + 182 = 0

⇒ x² -13x - 14x + 182 = 0

⇒ x(x - 13) - 14(x - 13) = 0

⇒ (x - 14)(x - 13) = 0

⇒ x - 14 = 0 or x - 13 = 0

⇒ x = 14 or x = 13.

Hence, numbers are 13 and 14.

Find two consecutive positive integers, sum of whose squares is 365.

Answer

Let two consecutive positive integers be x and (x + 1).

According to question,

⇒ x² + (x + 1)² = 365

⇒ x² + x² + 1 + 2x = 365

⇒ 2x² + 2x + 1 - 365 = 0

⇒ 2x² + 2x - 364 = 0

⇒ 2(x² + x - 182) = 0

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x(x + 14) - 13(x + 14) = 0

⇒ (x - 13)(x + 14) = 0

⇒ x - 13 = 0 or x + 14 = 0

⇒ x = 13 or x = -14.

Since, positive integer cannot be negative.

∴ x = 13 and x + 1 = 14.

Hence, required positive integers are 13 and 14.

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Answer

Given,

Hypotenuse = 13 cm

Let base = x cm, so perpendicular = (x - 7) cm.

By pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

⇒ 13² = (x - 7)² + x²

⇒ 169 = x² + 49 - 14x + x²

⇒ 169 = 2x² - 14x + 49

⇒ 2x² - 14x + 49 - 169 = 0

⇒ 2x² - 14x - 120 = 0

⇒ 2(x² - 7x - 60) = 0

⇒ x² - 7x - 60 = 0

⇒ x² - 12x + 5x - 60 = 0

⇒ x(x - 12) + 5(x - 12) = 0

⇒ (x + 5)(x - 12) = 0

⇒ x + 5 = 0 or x - 12 = 0

⇒ x = -5 or x = 12.

Since, side cannot be negative.

∴ Base(x) = 12 cm, Altitude (x - 7) = 5 cm.

Hence, other two sides are 5 cm and 12 cm.

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.

Answer

Let no. of articles produced be x.

So, cost of production of each article = (2x + 3)

Given,

Total cost = 90

⇒ x(2x + 3) = 90

⇒ 2x² + 3x = 90

⇒ 2x² + 3x - 90 = 0

⇒ 2x² + 15x - 12x - 90 = 0

⇒ x(2x + 15) - 6(2x + 15) = 0

⇒ (x - 6)(2x + 15) = 0

⇒ x - 6 = 0 or 2x + 15 = 0

⇒ x = 6 or x = $-\dfrac{15}{2}$.

Since, no. of articles cannot be negative.

∴ x = 6 and 2x + 3 = 2(6) + 3 = 12 + 3 = 15.

Hence, no. of articles produced = 6 and cost of each article = ₹ 15.

Find the nature of the roots of the following quadratic equation. If the real roots exist, find them :

2x² - 3x + 5 = 0

Answer

Comparing 2x² - 3x + 5 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -3 and c = 5.

Substituting values in b² - 4ac, we get :

⇒ (-3)² - 4(2)(5)

⇒ 9 - 40

⇒ -31.

Since, b² - 4ac < 0.

∴ There are no real roots.

Hence, real roots does not exist.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

3x² - $4\sqrt{3}x + 4$ = 0

Answer

Comparing 3x² - $4\sqrt{3}x + 4$ = 0 with ax² + bx + c = 0, we get :

a = 3, b = $-4\sqrt{3}$ and c = 4.

Substituting values in b² - 4ac, we get :

$\Rightarrow (-4\sqrt{3})^2 - 4 \times 3 \times 4 \\[1em] \Rightarrow 48 - 48 \\[1em] \Rightarrow 0.$

Since, b² - 4ac = 0.

∴ There are two real and equal roots.

For equal roots,

x = $-\dfrac{b}{2a} = -\dfrac{-4\sqrt{3}}{2 \times 3} = \dfrac{4\sqrt{3}}{6} = \dfrac{2\sqrt{3}}{3}$.

Hence, equal roots are $\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}$.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

2x² - 6x + 3 = 0

Answer

Comparing 2x² - 6x + 3 = 0 with ax² + bx + c = 0, we get :

a = 2, b = -6 and c = 3.

Substituting values in b² - 4ac, we get :

$\Rightarrow (-6)^2 - 4 \times 2 \times 3 \\[1em] \Rightarrow 36 - 24 \\[1em] \Rightarrow 12.$

Since, b² - 4ac > 0.

∴ There are two real and distinct roots.

Solving,

⇒ 2x² - 6x + 3 = 0

For distinct real roots,

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

Substituting values we get :

$\Rightarrow x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 2 \times 3}}{2 \times 2} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{36 - 24}}{4} \\[1em] \Rightarrow x = \dfrac{6 \pm \sqrt{12}}{4} \\[1em] \Rightarrow x = \dfrac{6 \pm 2\sqrt{3}}{4} \\[1em] \Rightarrow x = \dfrac{3 \pm \sqrt{3}}{2}.$

Hence, x = $\dfrac{3 \pm \sqrt{3}}{2}.$

Find the value of k for the following quadratic equation, so that it has two equal roots.

2x² + kx + 3 = 0

Answer

Comparing 2x² + kx + 3 = 0 with ax² + bx + c = 0, we get :

a = 2, b = k and c = 3.

For equal roots,

⇒ b² - 4ac = 0

⇒ k² - 4 × 2 × 3 = 0

⇒ k² - 24 = 0

⇒ k² = 24

⇒ k = $\sqrt{24}$

⇒ k = $\pm 2\sqrt{6}$

Hence, k = $\pm 2\sqrt{6}$.

Find the value of k for the following quadratic equation, so that it has two equal roots.

kx(x - 2) + 6 = 0

Answer

Given,

Equation : kx(x - 2) + 6 = 0

⇒ kx² - 2kx + 6 = 0

Comparing kx² - 2kx + 6 = 0 with ax² + bx + c = 0, we get :

a = k, b = -2k and c = 6.

For equal roots,

⇒ b² - 4ac = 0

⇒ (-2k)² - 4 × k × 6 = 0

⇒ 4k² - 24k = 0

⇒ 4k(k - 6) = 0

⇒ 4k = 0 or k - 6 = 0

⇒ k = 0 or k = 6.

k cannot be equal to zero because then equation kx² - 2kx + 6 = 0, will not be quadratic.

∴ k = 6.

Hence, 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.

Answer

Let breadth be x meters and so length is 2x meters.

By formula,

Area = Length × Breadth

⇒ 800 = 2x.x

⇒ 2x² = 800

⇒ x² = $\dfrac{800}{2}$

⇒ x² = 400

⇒ x² - 400 = 0.

Comparing x² - 400 = 0 with ax² + bx + c = 0, we get :

a = 1, b = 0 and c = -400.

Substituting value in b² - 4ac, we get :

⇒ 0² - 4 × 1 × -400

⇒ 0 + 1600

⇒ 1600.

Since, b² - 4ac > 0,

∴ Equation has real and distinct roots.

Solving, x² - 400 = 0, we get :

⇒ x² = 400

⇒ x = $\sqrt{400}$

⇒ x = ± 20

Since, side cannot be negative.

∴ x ≠ -20.

∴ Breadth (x) = 20 meters and Length (2x) = 2 × 20 = 40 meters.

Hence situation is possible and length = 40 meters and breadth = 20 meters.

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.

Answer

Let age of first friend be x years, so age of other friend is (20 - x) years.

Four years ago,

Age of first friend = (x - 4) years and age of second friend = (20 - x - 4) = (16 - x) years.

Given,

Product of their ages in years was 48.

⇒ (x - 4)(16 - x) = 48

⇒ 16x - x² - 64 + 4x = 48

⇒ 20x - x² - 64 = 48

⇒ x² - 20x + 64 + 48 = 0

⇒ x² - 20x + 112 = 0

Comparing x² - 20x + 112 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -20 and c = 112.

Substituting value in b² - 4ac, we get :

⇒ (-20)² - 4 × 1 × 112

⇒ 400 - 448

⇒ -48.

Since, b² - 4ac < 0.

∴ Roots are imaginary.

Hence, the above situation is not possible.

Is it possible to design a rectangular park of perimeter 80 m and area 400 m² ? If so, find its length and breadth.

Answer

Let length of rectangular park be l and breadth be b meters.

Given,

Perimeter = 80 meters.

⇒ 2(l + b) = 80

⇒ l + b = 40

⇒ b = 40 - l ...........(1)

Given,

⇒ Area = 400 m²

⇒ lb = 400

⇒ l(40 - l) = 400

⇒ 40l - l² = 400

⇒ l² - 40l + 400 = 0

Comparing l² - 40l + 400 = 0 with ax² + bx + c = 0, we get :

a = 1, b = -40 and c = 400.

Substituting value in b² - 4ac, we get :

⇒ (-40)² - 4 × 1 × 400

⇒ 1600 - 1600

⇒ 0.

Since, b² - 4ac = 0.

∴ Roots are real and equal.

Solving, l² - 40l + 400 = 0, we get :

⇒ l² - 20l - 20l + 400 = 0

⇒ l(l - 20) - 20(l - 20) = 0

⇒ (l - 20)(l - 20) = 0

⇒ l - 20 = 0

⇒ l = 20 meters.

Breadth (b) = 40 - l = 40 - 20 = 20 meters.

Hence, condition is possible and length = 20 meters and breadth = 20 meters.