Class 10 NCERT Solutions
Chapter 4: Quadratic Equations
Master the splitting of middle terms, the quadratic formula, and the logic of discriminants to determine the nature of roots with our step-by-step approach.
Exercise 4.1
1. 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) $x^3-4x^2-x+1=(x-2)^2$
(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) $x^3-4x^2-x+1=(x-2)^2$
(i) Start by expanding both sides:
$(x+1)^2=2(x-3) \Rightarrow x^2+2x+1=2x-6 \Rightarrow x^2+7=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(ii) Expand and simplify:
$x^2-2x=(-2)(3-x) \Rightarrow x^2-2x=-6+2x$
$\Rightarrow x^2-4x+6=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(iii) Expand both sides:
$(x-2)(x+1)=(x-1)(x+3)$
$\Rightarrow x^2-x-2=x^2+2x-3 \Rightarrow 3x-1=0$
The $x^2$ terms cancel out, so this does not match $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(iv) Expand and rearrange:
$(x-3)(2x+1)=x(x+5)$
$\Rightarrow 2x^2+x-6x-3=x^2+5x$
$\Rightarrow x^2-10x-3=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(v) Expand both sides and simplify:
$(2x-1)(x-3)=(x+5)(x-1)$
$\Rightarrow 2x^2-7x+3=x^2+4x-5$
$\Rightarrow x^2-11x+8=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(vi) Expand the right side:
$x^2+3x+1=(x-2)^2$
$\Rightarrow x^2+3x+1=x^2-4x+4$
$\Rightarrow 7x-3=0$
The $x^2$ terms cancel, so this does not match $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(vii) Expand both sides:
$(x+2)^3=2x(x^2-1)$
$\Rightarrow x^3+6x^2+12x+8=2x^3-2x$
$\Rightarrow x^3-6x^2-14x-8=0$
This is a cubic equation, not of the form $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(viii) Expand the right side and simplify:
$x^3-4x^2-x+1=(x-2)^3$
$\Rightarrow x^3-4x^2-x+1=x^3-6x^2+12x-8$
$\Rightarrow 2x^2-13x+9=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
$(x+1)^2=2(x-3) \Rightarrow x^2+2x+1=2x-6 \Rightarrow x^2+7=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(ii) Expand and simplify:
$x^2-2x=(-2)(3-x) \Rightarrow x^2-2x=-6+2x$
$\Rightarrow x^2-4x+6=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(iii) Expand both sides:
$(x-2)(x+1)=(x-1)(x+3)$
$\Rightarrow x^2-x-2=x^2+2x-3 \Rightarrow 3x-1=0$
The $x^2$ terms cancel out, so this does not match $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(iv) Expand and rearrange:
$(x-3)(2x+1)=x(x+5)$
$\Rightarrow 2x^2+x-6x-3=x^2+5x$
$\Rightarrow x^2-10x-3=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(v) Expand both sides and simplify:
$(2x-1)(x-3)=(x+5)(x-1)$
$\Rightarrow 2x^2-7x+3=x^2+4x-5$
$\Rightarrow x^2-11x+8=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
(vi) Expand the right side:
$x^2+3x+1=(x-2)^2$
$\Rightarrow x^2+3x+1=x^2-4x+4$
$\Rightarrow 7x-3=0$
The $x^2$ terms cancel, so this does not match $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(vii) Expand both sides:
$(x+2)^3=2x(x^2-1)$
$\Rightarrow x^3+6x^2+12x+8=2x^3-2x$
$\Rightarrow x^3-6x^2-14x-8=0$
This is a cubic equation, not of the form $ax^2+bx+c=0,\ a \neq 0$. Not a quadratic equation.
(viii) Expand the right side and simplify:
$x^3-4x^2-x+1=(x-2)^3$
$\Rightarrow x^3-4x^2-x+1=x^3-6x^2+12x-8$
$\Rightarrow 2x^2-13x+9=0$
This matches the standard form $ax^2+bx+c=0,\ a \neq 0$. It is a quadratic equation.
2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is $528\ \text{m}^2$. The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth.
(ii) The product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan’s present age.
(iv) A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Find the speed of the train.
(i) The area of a rectangular plot is $528\ \text{m}^2$. The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth.
(ii) The product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages 3 years from now will be 360. Find Rohan’s present age.
(iv) A train travels 480 km at a uniform speed. If the speed had been 8 km/h less, it would have taken 3 hours more. Find the speed of the train.
(i) Let the breadth of the plot $= x$ m. Then the length $= (2x+1)$ m.
Applying the area condition:
$(2x+1) \cdot x = 528$
$\Rightarrow 2x^2+x-528=0$ is the required quadratic equation.
(ii) Let the two consecutive positive integers be $x$ and $x+1$.
Applying the product condition:
$x(x+1)=306$
$\Rightarrow x^2+x-306=0$ is the required quadratic equation.
(iii) Let Rohan’s present age $= x$ years. Then his mother’s present age $= (x+26)$ years.
Three years from now: Rohan’s age $= (x+3)$ years and his mother’s age $= (x+29)$ years.
Applying the product condition:
$(x+3)(x+29)=360 \Rightarrow x^2+32x+87=360$
$\Rightarrow x^2+32x-273=0$ is the required quadratic equation.
(iv) Let the speed of the train $= x$ km/hr. Distance $= 480$ km.
Time taken at original speed $= \frac{480}{x}$ hours.
New speed $= (x-8)$ km/hr, and new time $= \frac{480}{x-8}$ hours.
Since the new journey takes 3 extra hours:
$\frac{480}{x-8}-\frac{480}{x}=3 \Rightarrow 480x – 480x + 3840 = 3x(x-8)$
$\Rightarrow x^2-8x-1280=0$ is the required quadratic equation.
Applying the area condition:
$(2x+1) \cdot x = 528$
$\Rightarrow 2x^2+x-528=0$ is the required quadratic equation.
(ii) Let the two consecutive positive integers be $x$ and $x+1$.
Applying the product condition:
$x(x+1)=306$
$\Rightarrow x^2+x-306=0$ is the required quadratic equation.
(iii) Let Rohan’s present age $= x$ years. Then his mother’s present age $= (x+26)$ years.
Three years from now: Rohan’s age $= (x+3)$ years and his mother’s age $= (x+29)$ years.
Applying the product condition:
$(x+3)(x+29)=360 \Rightarrow x^2+32x+87=360$
$\Rightarrow x^2+32x-273=0$ is the required quadratic equation.
(iv) Let the speed of the train $= x$ km/hr. Distance $= 480$ km.
Time taken at original speed $= \frac{480}{x}$ hours.
New speed $= (x-8)$ km/hr, and new time $= \frac{480}{x-8}$ hours.
Since the new journey takes 3 extra hours:
$\frac{480}{x-8}-\frac{480}{x}=3 \Rightarrow 480x – 480x + 3840 = 3x(x-8)$
$\Rightarrow x^2-8x-1280=0$ is the required quadratic equation.
Exercise 4.2
1. 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) $2x^2-x+\frac{1}{8}=0$ (v) $100x^2-20x+1=0$
(i) $x^2-3x-10=0$ (ii) $2x^2+x-6=0$ (iii) $\sqrt{2}x^2+7x+5\sqrt{2}=0$
(iv) $2x^2-x+\frac{1}{8}=0$ (v) $100x^2-20x+1=0$
(i) Split the middle term and factorise:
$x^2-3x-10=0$
$\Rightarrow x^2-5x+2x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow (x-5)(x+2)=0$
$\Rightarrow x=5$ or $x=-2$
(ii) Split the middle term and factorise:
$2x^2+x-6=0$
$\Rightarrow 2x^2+4x-3x-6=0$
$\Rightarrow 2x(x+2)-3(x+2)=0$
$\Rightarrow (2x-3)(x+2)=0$
$\Rightarrow x=\frac{3}{2}$ or $x=-2$
(iii) Split the middle term carefully:
$\sqrt{2}x^2+7x+5\sqrt{2}=0$
$\Rightarrow \sqrt{2}x^2+5x+2x+5\sqrt{2}=0$
$\Rightarrow x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$
$\Rightarrow (x+\sqrt{2})(\sqrt{2}x+5)=0$
$\Rightarrow x=-\sqrt{2}$ or $x=-\frac{5}{\sqrt{2}}$
(iv) Multiply through by 8 to clear the fraction first:
$2x^2-x+\frac{1}{8}=0 \Rightarrow 16x^2-8x+1=0$
$\Rightarrow 16x^2-4x-4x+1=0$
$\Rightarrow 4x(4x-1)-1(4x-1)=0$
$\Rightarrow (4x-1)(4x-1)=0$
$\Rightarrow x=\frac{1}{4},\ \frac{1}{4}$
(v) Split the middle term and factorise:
$100x^2-20x+1=0$
$\Rightarrow 100x^2-10x-10x+1=0$
$\Rightarrow 10x(10x-1)-1(10x-1)=0$
$\Rightarrow (10x-1)^2=0$
$\Rightarrow x=\frac{1}{10},\ \frac{1}{10}$
$x^2-3x-10=0$
$\Rightarrow x^2-5x+2x-10=0$
$\Rightarrow x(x-5)+2(x-5)=0$
$\Rightarrow (x-5)(x+2)=0$
$\Rightarrow x=5$ or $x=-2$
(ii) Split the middle term and factorise:
$2x^2+x-6=0$
$\Rightarrow 2x^2+4x-3x-6=0$
$\Rightarrow 2x(x+2)-3(x+2)=0$
$\Rightarrow (2x-3)(x+2)=0$
$\Rightarrow x=\frac{3}{2}$ or $x=-2$
(iii) Split the middle term carefully:
$\sqrt{2}x^2+7x+5\sqrt{2}=0$
$\Rightarrow \sqrt{2}x^2+5x+2x+5\sqrt{2}=0$
$\Rightarrow x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$
$\Rightarrow (x+\sqrt{2})(\sqrt{2}x+5)=0$
$\Rightarrow x=-\sqrt{2}$ or $x=-\frac{5}{\sqrt{2}}$
(iv) Multiply through by 8 to clear the fraction first:
$2x^2-x+\frac{1}{8}=0 \Rightarrow 16x^2-8x+1=0$
$\Rightarrow 16x^2-4x-4x+1=0$
$\Rightarrow 4x(4x-1)-1(4x-1)=0$
$\Rightarrow (4x-1)(4x-1)=0$
$\Rightarrow x=\frac{1}{4},\ \frac{1}{4}$
(v) Split the middle term and factorise:
$100x^2-20x+1=0$
$\Rightarrow 100x^2-10x-10x+1=0$
$\Rightarrow 10x(10x-1)-1(10x-1)=0$
$\Rightarrow (10x-1)^2=0$
$\Rightarrow x=\frac{1}{10},\ \frac{1}{10}$
2. Solve the problems given below:
(i) John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the number of marbles they now have is 124. Find the number of marbles each 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. Find the number of toys produced on that day.
(i) John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the number of marbles they now have is 124. Find the number of marbles each 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. Find the number of toys produced on that day.
(i) Let one of them start with $x$ marbles, so the other had $(45-x)$ marbles.
After losing 5 each:
$(x-5)(45-x-5)=124$
$\Rightarrow (x-5)(40-x)=124$
$\Rightarrow x^2-45x+324=0$
$\Rightarrow x^2-9x-36x+324=0$
$\Rightarrow x(x-9)-36(x-9)=0$
$\Rightarrow (x-9)(x-36)=0$
$\Rightarrow x=9$ or $x=36$
So if John started with 9 marbles, Jivanti had 36, and vice versa.
(ii) Let the number of toys produced $= x$. Then cost per toy $= ₹\frac{750}{x}$.
Applying the given condition:
$\frac{750}{x}=55-x \Rightarrow 750=55x-x^2$
$\Rightarrow x^2-55x+750=0$
$\Rightarrow x^2-25x-30x+750=0$
$\Rightarrow x(x-25)-30(x-25)=0$
$\Rightarrow (x-30)(x-25)=0$
$\Rightarrow x=30$ or $x=25$
The number of toys produced is 30 or 25.
After losing 5 each:
$(x-5)(45-x-5)=124$
$\Rightarrow (x-5)(40-x)=124$
$\Rightarrow x^2-45x+324=0$
$\Rightarrow x^2-9x-36x+324=0$
$\Rightarrow x(x-9)-36(x-9)=0$
$\Rightarrow (x-9)(x-36)=0$
$\Rightarrow x=9$ or $x=36$
So if John started with 9 marbles, Jivanti had 36, and vice versa.
(ii) Let the number of toys produced $= x$. Then cost per toy $= ₹\frac{750}{x}$.
Applying the given condition:
$\frac{750}{x}=55-x \Rightarrow 750=55x-x^2$
$\Rightarrow x^2-55x+750=0$
$\Rightarrow x^2-25x-30x+750=0$
$\Rightarrow x(x-25)-30(x-25)=0$
$\Rightarrow (x-30)(x-25)=0$
$\Rightarrow x=30$ or $x=25$
The number of toys produced is 30 or 25.
3. Find two numbers whose sum is 27 and product is 182.
Let the two numbers be $x$ and $y$.
From the given conditions: $x+y=27\ \ldots(i)$ and $xy=182\ \ldots(ii)$
Using (i) in (ii): $x(27-x)=182$
$\Rightarrow x^2-27x+182=0$
$\Rightarrow x^2-14x-13x+182=0$
$\Rightarrow x(x-14)-13(x-14)=0$
$\Rightarrow (x-13)(x-14)=0$
$\Rightarrow x=13$ or $x=14$
When $x=13,\ y=14$ and when $x=14,\ y=13$.
The two numbers are 13 and 14.
From the given conditions: $x+y=27\ \ldots(i)$ and $xy=182\ \ldots(ii)$
Using (i) in (ii): $x(27-x)=182$
$\Rightarrow x^2-27x+182=0$
$\Rightarrow x^2-14x-13x+182=0$
$\Rightarrow x(x-14)-13(x-14)=0$
$\Rightarrow (x-13)(x-14)=0$
$\Rightarrow x=13$ or $x=14$
When $x=13,\ y=14$ and when $x=14,\ y=13$.
The two numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Let the two consecutive positive integers be $x$ and $x+1$.
Applying the given condition:
$x^2+(x+1)^2=365 \Rightarrow 2x^2+2x-364=0$
$\Rightarrow x^2+x-182=0$
$\Rightarrow x^2+14x-13x-182=0$
$\Rightarrow x(x+14)-13(x+14)=0$
$\Rightarrow (x+14)(x-13)=0$
$\Rightarrow x=-14$ (rejected since we need a positive integer) or $x=13$
The two consecutive positive integers are 13 and 14.
Applying the given condition:
$x^2+(x+1)^2=365 \Rightarrow 2x^2+2x-364=0$
$\Rightarrow x^2+x-182=0$
$\Rightarrow x^2+14x-13x-182=0$
$\Rightarrow x(x+14)-13(x+14)=0$
$\Rightarrow (x+14)(x-13)=0$
$\Rightarrow x=-14$ (rejected since we need a positive integer) or $x=13$
The two consecutive positive integers are 13 and 14.
5. 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 $= x$ cm. Then the altitude $= (x-7)$ cm, and the hypotenuse $= 13$ cm.
Applying the Pythagorean theorem:
$x^2+(x-7)^2=(13)^2$
$\Rightarrow x^2+x^2-14x+49=169$
$\Rightarrow 2x^2-14x-120=0$
$\Rightarrow x^2-7x-60=0$
$\Rightarrow (x-12)(x+5)=0$
$\Rightarrow x=12$ or $x=-5$ (rejected since length must be positive)
$\therefore$ Base $= 12$ cm and Altitude $= 5$ cm.
Applying the Pythagorean theorem:
$x^2+(x-7)^2=(13)^2$
$\Rightarrow x^2+x^2-14x+49=169$
$\Rightarrow 2x^2-14x-120=0$
$\Rightarrow x^2-7x-60=0$
$\Rightarrow (x-12)(x+5)=0$
$\Rightarrow x=12$ or $x=-5$ (rejected since length must be positive)
$\therefore$ Base $= 12$ cm and Altitude $= 5$ cm.
6. A cottage industry produces a certain number of pottery articles in a day. The cost of production of each article (in rupees) was 3 more than twice the number of articles produced. If the total cost was ₹90, find the number of articles produced and the cost of each article.
Let the number of articles produced $= x$. Then cost per article $= (2x+3)$ rupees.
Applying the total cost condition:
$(2x+3) \cdot x=90$
$\Rightarrow 2x^2+3x-90=0$
$\Rightarrow 2x^2+15x-12x-90=0$
$\Rightarrow x(2x+15)-6(2x+15)=0$
$\Rightarrow (2x+15)(x-6)=0$
$\Rightarrow x=-\frac{15}{2}$ (rejected) or $x=6$
Number of articles produced $= 6$ and cost of each article $= ₹15$.
Applying the total cost condition:
$(2x+3) \cdot x=90$
$\Rightarrow 2x^2+3x-90=0$
$\Rightarrow 2x^2+15x-12x-90=0$
$\Rightarrow x(2x+15)-6(2x+15)=0$
$\Rightarrow (2x+15)(x-6)=0$
$\Rightarrow x=-\frac{15}{2}$ (rejected) or $x=6$
Number of articles produced $= 6$ and cost of each article $= ₹15$.
Exercise 4.3
1. 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) $2x^2-3x+5=0$ (ii) $3x^2-4\sqrt{3}x+4=0$ (iii) $2x^2-6x+3=0$
(i) Here $a=2,\ b=-3,\ c=5$.
$\mathrm{D}=b^2-4ac=(-3)^2-4\times2\times5=9-40=-31<0$
Since $\mathrm{D}<0$, the equation has no real roots.
(ii) Here $a=3,\ b=-4\sqrt{3},\ c=4$.
$\mathrm{D}=b^2-4ac=(-4\sqrt{3})^2-4\times3\times4=48-48=0$
Since $\mathrm{D}=0$, the roots are real and equal.
$\alpha=\frac{-b+\sqrt{\mathrm{D}}}{2a}=\frac{4\sqrt{3}+0}{6}=\frac{2\sqrt{3}}{3}$ and $\beta=\frac{-b-\sqrt{\mathrm{D}}}{2a}=\frac{4\sqrt{3}-0}{6}=\frac{2\sqrt{3}}{3}$
Both roots are $\frac{2\sqrt{3}}{3}$.
(iii) Here $a=2,\ b=-6,\ c=3$.
$\mathrm{D}=b^2-4ac=(-6)^2-4\times2\times3=36-24=12>0$
Since $\mathrm{D}>0$, the roots are real and unequal.
$x=\frac{-(-6)\pm\sqrt{12}}{4}=\frac{6\pm2\sqrt{3}}{4}=\frac{3\pm\sqrt{3}}{2}$
The roots are $\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.
$\mathrm{D}=b^2-4ac=(-3)^2-4\times2\times5=9-40=-31<0$
Since $\mathrm{D}<0$, the equation has no real roots.
(ii) Here $a=3,\ b=-4\sqrt{3},\ c=4$.
$\mathrm{D}=b^2-4ac=(-4\sqrt{3})^2-4\times3\times4=48-48=0$
Since $\mathrm{D}=0$, the roots are real and equal.
$\alpha=\frac{-b+\sqrt{\mathrm{D}}}{2a}=\frac{4\sqrt{3}+0}{6}=\frac{2\sqrt{3}}{3}$ and $\beta=\frac{-b-\sqrt{\mathrm{D}}}{2a}=\frac{4\sqrt{3}-0}{6}=\frac{2\sqrt{3}}{3}$
Both roots are $\frac{2\sqrt{3}}{3}$.
(iii) Here $a=2,\ b=-6,\ c=3$.
$\mathrm{D}=b^2-4ac=(-6)^2-4\times2\times3=36-24=12>0$
Since $\mathrm{D}>0$, the roots are real and unequal.
$x=\frac{-(-6)\pm\sqrt{12}}{4}=\frac{6\pm2\sqrt{3}}{4}=\frac{3\pm\sqrt{3}}{2}$
The roots are $\frac{3+\sqrt{3}}{2}$ and $\frac{3-\sqrt{3}}{2}$.
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) $2x^2+kx+3=0$ (ii) $kx(x-2)+6=0$
(i) Here $a=2,\ b=k,\ c=3$.
$\mathrm{D}=b^2-4ac=k^2-4\times2\times3=k^2-24$
For equal roots, $\mathrm{D}=0$:
$k^2-24=0 \Rightarrow k^2=24 \Rightarrow k=\pm\sqrt{24} \Rightarrow k=\pm2\sqrt{6}$
(ii) Rewrite the equation: $kx(x-2)+6=0 \Rightarrow kx^2-2kx+6=0$
Here $a=k,\ b=-2k,\ c=6$.
$\mathrm{D}=(-2k)^2-4\times k\times6=4k^2-24k$
For equal roots, $\mathrm{D}=0$:
$4k^2-24k=0 \Rightarrow 4k(k-6)=0$
$\Rightarrow k=0$ or $k=6$
Since $k=0$ would make the equation invalid, $\therefore k=6$.
$\mathrm{D}=b^2-4ac=k^2-4\times2\times3=k^2-24$
For equal roots, $\mathrm{D}=0$:
$k^2-24=0 \Rightarrow k^2=24 \Rightarrow k=\pm\sqrt{24} \Rightarrow k=\pm2\sqrt{6}$
(ii) Rewrite the equation: $kx(x-2)+6=0 \Rightarrow kx^2-2kx+6=0$
Here $a=k,\ b=-2k,\ c=6$.
$\mathrm{D}=(-2k)^2-4\times k\times6=4k^2-24k$
For equal roots, $\mathrm{D}=0$:
$4k^2-24k=0 \Rightarrow 4k(k-6)=0$
$\Rightarrow k=0$ or $k=6$
Since $k=0$ would make the equation invalid, $\therefore k=6$.
3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is $800\ \text{m}^2$? If so, find its length and breadth.
Let the breadth $= x$ m, so the length $= 2x$ m.
Applying the area condition:
$x \times 2x=800 \Rightarrow 2x^2=800 \Rightarrow x^2=400$
$\Rightarrow x=20$ (we reject $-20$ since length must be positive)
Yes, it is possible. The rectangular mango grove has a length of 40 m and breadth of 20 m.
Applying the area condition:
$x \times 2x=800 \Rightarrow 2x^2=800 \Rightarrow x^2=400$
$\Rightarrow x=20$ (we reject $-20$ since length must be positive)
Yes, it is possible. The rectangular mango grove has a length of 40 m and breadth of 20 m.
4. 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 $= x$ years and the other $= y$ years.
From the first condition: $x+y=20\ \ldots(i)$
Four years ago: $(x-4)(y-4)=48$
$\Rightarrow xy-4x-4y+16=48$
$\Rightarrow xy-4(x+y)-32=0$
$\Rightarrow xy-4(20)-32=0 \Rightarrow xy=112\ \ldots(ii)$
From (i) and (ii): $x(20-x)=112 \Rightarrow x^2-20x+112=0$
$\mathrm{D}=(-20)^2-4\times1\times112=400-448=-48<0$
Since $\mathrm{D}<0$, there are no real roots. The situation is not possible.
From the first condition: $x+y=20\ \ldots(i)$
Four years ago: $(x-4)(y-4)=48$
$\Rightarrow xy-4x-4y+16=48$
$\Rightarrow xy-4(x+y)-32=0$
$\Rightarrow xy-4(20)-32=0 \Rightarrow xy=112\ \ldots(ii)$
From (i) and (ii): $x(20-x)=112 \Rightarrow x^2-20x+112=0$
$\mathrm{D}=(-20)^2-4\times1\times112=400-448=-48<0$
Since $\mathrm{D}<0$, there are no real roots. The situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80 m and area $400\ \text{m}^2$? If so, find its length and breadth.
Let length $= x$ m and breadth $= y$ m.
From the perimeter condition: $2(x+y)=80 \Rightarrow x+y=40\ \ldots(i)$
From the area condition: $xy=400\ \ldots(ii)$
Substituting from (i) into (ii): $x(40-x)=400 \Rightarrow x^2-40x+400=0$
$\mathrm{D}=(-40)^2-4\times1\times400=1600-1600=0$
Since $\mathrm{D}=0$, the equation has real and equal roots.
$\alpha=\frac{-(-40)+\sqrt{0}}{2\times1}=\frac{40}{2}=20$ and $\beta=\frac{-(-40)-\sqrt{0}}{2\times1}=\frac{40}{2}=20$
Yes, it is possible. The length and breadth are both equal to 20 m — the park is actually a square.
From the perimeter condition: $2(x+y)=80 \Rightarrow x+y=40\ \ldots(i)$
From the area condition: $xy=400\ \ldots(ii)$
Substituting from (i) into (ii): $x(40-x)=400 \Rightarrow x^2-40x+400=0$
$\mathrm{D}=(-40)^2-4\times1\times400=1600-1600=0$
Since $\mathrm{D}=0$, the equation has real and equal roots.
$\alpha=\frac{-(-40)+\sqrt{0}}{2\times1}=\frac{40}{2}=20$ and $\beta=\frac{-(-40)-\sqrt{0}}{2\times1}=\frac{40}{2}=20$
Yes, it is possible. The length and breadth are both equal to 20 m — the park is actually a square.
