Quadratic Equations — CBSE Class 10 Maths Textbook Solutions

(i) LHS $=(x+1)^2=x^2+2x+1$=(x+1)^2=x^2+2x+1. Equation becomes $x^2+2x+1=2x-6 \Rightarrow x^2+7=0$x^2+2x+1=2x-6 x^2+7=0. This is of the form $ax^2+bx+c=0$ax^2+bx+c=0 (with $a=1,b=0,c=7$a=1,b=0,c=7). It is a quadratic equation.

(ii) RHS $=(-2)(3-x)=-6+2x$=(-2)(3-x)=-6+2x. Equation becomes $x^2-2x=-6+2x \Rightarrow x^2-4x+6=0$x^2-2x=-6+2x x^2-4x+6=0. This is of the required form. It is a quadratic equation.

(iii) LHS $=(x-2)(x+1)=x^2-x-2$=(x-2)(x+1)=x^2-x-2. RHS $=(x-1)(x+3)=x^2+2x-3$=(x-1)(x+3)=x^2+2x-3. Equation becomes $x^2-x-2=x^2+2x-3 \Rightarrow -3x+1=0$x^2-x-2=x^2+2x-3 -3x+1=0, i.e. $3x-1=0$3x-1=0, a linear equation. It is not a quadratic equation.

(iv) LHS $=(x-3)(2x+1)=2x^2-5x-3$=(x-3)(2x+1)=2x^2-5x-3. RHS $=x(x+5)=x^2+5x$=x(x+5)=x^2+5x. Equation becomes $2x^2-5x-3=x^2+5x \Rightarrow x^2-10x-3=0$2x^2-5x-3=x^2+5x x^2-10x-3=0. This is of the required form. It is a quadratic equation.

(v) LHS $=(2x-1)(x-3)=2x^2-7x+3$=(2x-1)(x-3)=2x^2-7x+3. RHS $=(x+5)(x-1)=x^2+4x-5$=(x+5)(x-1)=x^2+4x-5. Equation becomes $2x^2-7x+3=x^2+4x-5 \Rightarrow x^2-11x+8=0$2x^2-7x+3=x^2+4x-5 x^2-11x+8=0. This is of the required form. It is a quadratic equation.

(vi) RHS $=(x-2)^2=x^2-4x+4$=(x-2)^2=x^2-4x+4. Equation becomes $x^2+3x+1=x^2-4x+4 \Rightarrow 7x-3=0$x^2+3x+1=x^2-4x+4 7x-3=0, a linear equation. It is not a quadratic equation.

(vii) LHS $=(x+2)^3=x^3+6x^2+12x+8$=(x+2)^3=x^3+6x^2+12x+8. RHS $=2x(x^2-1)=2x^3-2x$=2x(x^2-1)=2x^3-2x. Equation becomes $x^3+6x^2+12x+8=2x^3-2x \Rightarrow x^3-6x^2-14x-8=0$x^3+6x^2+12x+8=2x^3-2x x^3-6x^2-14x-8=0, which has an $x^3$x^3 term — it is a cubic equation. It is not a quadratic equation.

(viii) RHS $=(x-2)^3=x^3-6x^2+12x-8$=(x-2)^3=x^3-6x^2+12x-8. Equation becomes $x^3-4x^2-x+1=x^3-6x^2+12x-8$x^3-4x^2-x+1=x^3-6x^2+12x-8. The $x^3$x^3 terms cancel: $-4x^2-x+1=-6x^2+12x-8 \Rightarrow 2x^2-13x+9=0$-4x^2-x+1=-6x^2+12x-8 2x^2-13x+9=0. This is of the required form. It is a quadratic equation.

(i) Let the breadth of the plot be $x$x metres. Then the length is $(2x+1)$(2x+1) metres.

Area $=$= length $\times$× breadth: $x(2x+1)=528 \Rightarrow 2x^2+x-528=0$x(2x+1)=528 2x^2+x-528=0

Required equation: $2x^2+x-528=0$2x^2+x-528=0, where $x$x is the breadth in metres.

(ii) Let the smaller of the two consecutive positive integers be $x$x. Then the next integer is $x+1$x+1.

$x(x+1)=306 \Rightarrow x^2+x-306=0$x(x+1)=306 x^2+x-306=0

Required equation: $x^2+x-306=0$x^2+x-306=0, where $x$x is the smaller integer.

(iii) Let Rohan's present age be $x$x years. His mother's present age is $(x+26)$(x+26) years.

Three years from now: Rohan's age $=x+3$=x+3, mother's age $=x+29$=x+29.

$(x+3)(x+29)=360 \Rightarrow x^2+29x+3x+87=360 \Rightarrow x^2+32x+87-360=0 \Rightarrow x^2+32x-273=0$(x+3)(x+29)=360 x^2+29x+3x+87=360 x^2+32x+87-360=0 x^2+32x-273=0

Required equation: $x^2+32x-273=0$x^2+32x-273=0, where $x$x is Rohan's present age in years.

(iv) Let the uniform speed of the train be $u$u km/h. Time taken to cover 480 km $=\dfrac{480}{u}$=480/u hours.

At speed $(u-8)$(u-8) km/h, time taken $=\dfrac{480}{u-8}$=480/u-8 hours, and this is 3 hours more:

$\dfrac{480}{u-8}-\dfrac{480}{u}=3$480/u-8-480/u=3

Multiply throughout by $u(u-8)$u(u-8): $480u-480(u-8)=3u(u-8) \Rightarrow 480u-480u+3840=3u^2-24u \Rightarrow 3840=3u^2-24u$480u-480(u-8)=3u(u-8) 480u-480u+3840=3u^2-24u 3840=3u^2-24u

Dividing by 3: $1280=u^2-8u \Rightarrow u^2-8u-1280=0$1280=u^2-8u u^2-8u-1280=0

Required equation: $u^2-8u-1280=0$u^2-8u-1280=0, where $u$u is the speed of the train in km/h.

(i) $x^2-3x-10=0$x^2-3x-10=0. Split $-3x$-3x into $-5x+2x$-5x+2x (since $-5 \times 2=-10$-5 × 2=-10 and $-5+2=-3$-5+2=-3):

$x^2-5x+2x-10=0 \Rightarrow x(x-5)+2(x-5)=0 \Rightarrow (x-5)(x+2)=0$x^2-5x+2x-10=0 x(x-5)+2(x-5)=0 (x-5)(x+2)=0

So $x-5=0$x-5=0 or $x+2=0$x+2=0, giving $x=5$x=5 or $x=-2$x=-2.

(ii) $2x^2+x-6=0$2x^2+x-6=0. Split $x$x into $4x-3x$4x-3x (since $4\times(-3)=-12=2\times(-6)$4×(-3)=-12=2×(-6) and $4+(-3)=1$4+(-3)=1):

$2x^2+4x-3x-6=0 \Rightarrow 2x(x+2)-3(x+2)=0 \Rightarrow (2x-3)(x+2)=0$2x^2+4x-3x-6=0 2x(x+2)-3(x+2)=0 (2x-3)(x+2)=0

So $2x-3=0$2x-3=0 or $x+2=0$x+2=0, giving $x=\dfrac{3}{2}$x=3/2 or $x=-2$x=-2.

(iii) $\sqrt{2}x^2+7x+5\sqrt{2}=0$√2x^2+7x+5√2=0. We need two numbers with product $\sqrt{2}\times 5\sqrt{2}=10$√2× 5√2=10 and sum $7$7: these are $2$2 and $5$5.

$\sqrt{2}x^2+2x+5x+5\sqrt{2}=0 \Rightarrow \sqrt{2}x(x+\sqrt{2})+5(x+\sqrt{2})=0 \Rightarrow (\sqrt{2}x+5)(x+\sqrt{2})=0$√2x^2+2x+5x+5√2=0 √2x(x+√2)+5(x+√2)=0 (√2x+5)(x+√2)=0

So $\sqrt{2}x+5=0$√2x+5=0 or $x+\sqrt{2}=0$x+√2=0, giving $x=-\dfrac{5}{\sqrt{2}}$x=-5/√2 or $x=-\sqrt{2}$x=-√2.

(iv) $2x^2-x+\dfrac{1}{8}=0$2x^2-x+1/8=0. Multiply throughout by 8: $16x^2-8x+1=0$16x^2-8x+1=0.

This is a perfect square: $16x^2-8x+1=(4x-1)^2=0 \Rightarrow 4x-1=0 \Rightarrow x=\dfrac{1}{4}$16x^2-8x+1=(4x-1)^2=0 4x-1=0 x=1/4.

Repeated root: $x=\dfrac{1}{4}, \dfrac{1}{4}$x=1/4, 1/4.

(v) $100x^2-20x+1=0$100x^2-20x+1=0. This is a perfect square: $100x^2-20x+1=(10x-1)^2=0 \Rightarrow 10x-1=0 \Rightarrow x=\dfrac{1}{10}$100x^2-20x+1=(10x-1)^2=0 10x-1=0 x=1/10.

Repeated root: $x=\dfrac{1}{10}, \dfrac{1}{10}$x=1/10, 1/10.

(i) From Example 1, if John had $x$x marbles, Jivanti had $(45-x)$(45-x) marbles, and after each lost 5 marbles, their product being 124 leads to the equation:

$x^2-45x+324=0$x^2-45x+324=0

Split $-45x$-45x into $-36x-9x$-36x-9x (since $-36\times(-9)=324$-36×(-9)=324 and $-36+(-9)=-45$-36+(-9)=-45):

$x^2-36x-9x+324=0 \Rightarrow x(x-36)-9(x-36)=0 \Rightarrow (x-36)(x-9)=0$x^2-36x-9x+324=0 x(x-36)-9(x-36)=0 (x-36)(x-9)=0

So $x=36$x=36 or $x=9$x=9.

If John had 36 marbles, Jivanti had $45-36=9$45-36=9 marbles. If John had 9 marbles, Jivanti had $45-9=36$45-9=36 marbles. Both cases are valid (a positive number of marbles left after losing 5 in each case: $36-5=31>0$36-5=31>0 and $9-5=4>0$9-5=4>0).

John and Jivanti had 36 and 9 marbles between them (in either order).

(ii) From Example 1, if $x$x toys were produced that day, the equation is:

$x^2-55x+750=0$x^2-55x+750=0

Split $-55x$-55x into $-25x-30x$-25x-30x (since $-25\times(-30)=750$-25×(-30)=750 and $-25+(-30)=-55$-25+(-30)=-55):

$x^2-25x-30x+750=0 \Rightarrow x(x-25)-30(x-25)=0 \Rightarrow (x-25)(x-30)=0$x^2-25x-30x+750=0 x(x-25)-30(x-25)=0 (x-25)(x-30)=0

So $x=25$x=25 or $x=30$x=30.

The number of toys produced that day was 25 or 30 (both give a valid positive cost per toy: ₹30 and ₹25 respectively).

Let the two numbers be $x$x and $27-x$27-x.

$x(27-x)=182 \Rightarrow 27x-x^2=182 \Rightarrow x^2-27x+182=0$x(27-x)=182 27x-x^2=182 x^2-27x+182=0

Split $-27x$-27x into $-13x-14x$-13x-14x (since $-13\times(-14)=182$-13×(-14)=182 and $-13+(-14)=-27$-13+(-14)=-27):

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

So $x=13$x=13 or $x=14$x=14.

The two numbers are 13 and 14. (Check: $13+14=27$13+14=27 and $13\times14=182$13×14=182.)

Let the two consecutive positive integers be $x$x and $x+1$x+1.

$x^2+(x+1)^2=365 \Rightarrow x^2+x^2+2x+1=365 \Rightarrow 2x^2+2x-364=0$x^2+(x+1)^2=365 x^2+x^2+2x+1=365 2x^2+2x-364=0

Dividing by 2: $x^2+x-182=0$x^2+x-182=0

Split $x$x into $14x-13x$14x-13x (since $14\times(-13)=-182$14×(-13)=-182 and $14+(-13)=1$14+(-13)=1):

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

So $x=-14$x=-14 or $x=13$x=13. Since the integers must be positive, we reject $x=-14$x=-14 and take $x=13$x=13.

The two consecutive positive integers are 13 and 14. (Check: $13^2+14^2=169+196=365$13^2+14^2=169+196=365.)

Let the base of the right triangle be $x$x cm. Then the altitude is $(x-7)$(x-7) cm.

By the Pythagoras theorem: (base)$^2$^2 + (altitude)$^2$^2 = (hypotenuse)$^2$^2

$x^2+(x-7)^2=13^2 \Rightarrow x^2+x^2-14x+49=169 \Rightarrow 2x^2-14x-120=0$x^2+(x-7)^2=13^2 x^2+x^2-14x+49=169 2x^2-14x-120=0

Dividing by 2: $x^2-7x-60=0$x^2-7x-60=0

Split $-7x$-7x into $-12x+5x$-12x+5x (since $-12\times5=-60$-12×5=-60 and $-12+5=-7$-12+5=-7):

$x^2-12x+5x-60=0 \Rightarrow x(x-12)+5(x-12)=0 \Rightarrow (x-12)(x+5)=0$x^2-12x+5x-60=0 x(x-12)+5(x-12)=0 (x-12)(x+5)=0

So $x=12$x=12 or $x=-5$x=-5. Since a side length cannot be negative, $x=-5$x=-5 is rejected, so $x=12$x=12.

The base is 12 cm and the altitude is $12-7=5$12-7=5 cm.

The other two sides are 5 cm and 12 cm. (Check: $5^2+12^2=25+144=169=13^2$5^2+12^2=25+144=169=13^2.)

Let the number of articles produced that day be $n$n. The cost of production of each article is $(2n+3)$(2n+3) rupees.

Total cost $=$= (number of articles) $\times$× (cost of each article):

$n(2n+3)=90 \Rightarrow 2n^2+3n-90=0$n(2n+3)=90 2n^2+3n-90=0

Split $3n$3n into $-12n+15n$-12n+15n (since $-12\times15=-180=2\times(-90)$-12×15=-180=2×(-90) and $-12+15=3$-12+15=3):

$2n^2-12n+15n-90=0 \Rightarrow 2n(n-6)+15(n-6)=0 \Rightarrow (2n+15)(n-6)=0$2n^2-12n+15n-90=0 2n(n-6)+15(n-6)=0 (2n+15)(n-6)=0

So $n=6$n=6 or $n=-\dfrac{15}{2}$n=-15/2. Since the number of articles cannot be negative or fractional, $n=6$n=6.

Cost of each article $=2(6)+3=15$=2(6)+3=15.

Number of articles produced = 6, cost of each article = ₹15. (Check: $6\times15=90$6×15=90.)

(i) Here $a=2,\ b=-3,\ c=5$a=2, b=-3, c=5. Discriminant $D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31$D=b^2-4ac=(-3)^2-4(2)(5)=9-40=-31.

Since $D<0$D<0, the equation has no real roots.

(ii) Here $a=3,\ b=-4\sqrt{3},\ c=4$a=3, b=-4√3, c=4. Discriminant $D=b^2-4ac=(-4\sqrt{3})^2-4(3)(4)=48-48=0$D=b^2-4ac=(-4√3)^2-4(3)(4)=48-48=0.

Since $D=0$D=0, the equation has two equal real roots, given by $x=-\dfrac{b}{2a}=\dfrac{4\sqrt{3}}{6}=\dfrac{2\sqrt{3}}{3}=\dfrac{2}{\sqrt{3}}$x=-b/2a=4√3/6=2√3/3=2/√3.

Roots: $x=\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}$x=2/√3, 2/√3.

(iii) Here $a=2,\ b=-6,\ c=3$a=2, b=-6, c=3. Discriminant $D=b^2-4ac=(-6)^2-4(2)(3)=36-24=12$D=b^2-4ac=(-6)^2-4(2)(3)=36-24=12.

Since $D>0$D>0, the equation has two distinct real roots, given by:

$x=\dfrac{-b\pm\sqrt{D}}{2a}=\dfrac{6\pm\sqrt{12}}{4}=\dfrac{6\pm2\sqrt{3}}{4}=\dfrac{3\pm\sqrt{3}}{2}$x=-b±√D/2a=6±√12/4=6±2√3/4=3±√3/2

Roots: $x=\dfrac{3+\sqrt{3}}{2}$x=3+√3/2 and $x=\dfrac{3-\sqrt{3}}{2}$x=3-√3/2.

(i) Here $a=2,\ b=k,\ c=3$a=2, b=k, c=3. For two equal roots, the discriminant must be zero:

$D=b^2-4ac=k^2-4(2)(3)=k^2-24=0 \Rightarrow k^2=24 \Rightarrow k=\pm2\sqrt{6}$D=b^2-4ac=k^2-4(2)(3)=k^2-24=0 k^2=24 k=±2√6

$k=2\sqrt{6}$k=2√6 or $k=-2\sqrt{6}$k=-2√6.

(ii) $kx(x-2)+6=0 \Rightarrow kx^2-2kx+6=0$kx(x-2)+6=0 kx^2-2kx+6=0. Here $a=k,\ b=-2k,\ c=6$a=k, b=-2k, c=6. For two equal roots:

$D=b^2-4ac=(-2k)^2-4(k)(6)=4k^2-24k=0 \Rightarrow 4k(k-6)=0 \Rightarrow k=0 \text{ or } k=6$D=b^2-4ac=(-2k)^2-4(k)(6)=4k^2-24k=0 4k(k-6)=0 k=0 or k=6

If $k=0$k=0, the equation is no longer quadratic (the $x^2$x^2 term vanishes), so this value is rejected.

$k=6$k=6.

Let the breadth of the grove be $x$x m. Then the length is $2x$2x m.

Area $=$= length $\times$× breadth: $2x \times x = 800 \Rightarrow 2x^2=800 \Rightarrow x^2=400 \Rightarrow x^2-400=0$2x × x = 800 2x^2=800 x^2=400 x^2-400=0

This is of the form $ax^2+bx+c=0$ax^2+bx+c=0 with $a=1, b=0, c=-400$a=1, b=0, c=-400. Discriminant $D=b^2-4ac=0-4(1)(-400)=1600>0$D=b^2-4ac=0-4(1)(-400)=1600>0.

Since $D>0$D>0, real roots exist, so it is possible.

Solving: $x^2=400 \Rightarrow x=\pm20$x^2=400 x=±20. Since breadth cannot be negative, $x=20$x=20.

Breadth = 20 m, Length $=2(20)=40$=2(20)=40 m.

Let the present age of one friend be $x$x years. Since their ages add up to 20, the other friend's present age is $(20-x)$(20-x) years.

Four years ago, their ages were $(x-4)$(x-4) and $(16-x)$(16-x) years respectively, and their product was 48:

$(x-4)(16-x)=48 \Rightarrow 16x-x^2-64+4x=48 \Rightarrow -x^2+20x-64=48 \Rightarrow -x^2+20x-112=0$(x-4)(16-x)=48 16x-x^2-64+4x=48 -x^2+20x-64=48 -x^2+20x-112=0

Multiplying by $-1$-1: $x^2-20x+112=0$x^2-20x+112=0

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

Since $D<0$D<0, there are no real roots, so this situation is not possible — two friends whose present ages sum to 20 years cannot have had a product of ages equal to 48 four years ago.

Let the length of the park be $x$x m. Since the perimeter is 80 m, $2(\text{length}+\text{breadth})=80 \Rightarrow \text{length}+\text{breadth}=40$2(length+breadth)=80 length+breadth=40, so the breadth is $(40-x)$(40-x) m.

Area $=$= length $\times$× breadth: $x(40-x)=400 \Rightarrow 40x-x^2=400 \Rightarrow x^2-40x+400=0$x(40-x)=400 40x-x^2=400 x^2-40x+400=0

Discriminant $D=b^2-4ac=(-40)^2-4(1)(400)=1600-1600=0$D=b^2-4ac=(-40)^2-4(1)(400)=1600-1600=0

Since $D=0$D=0, real (equal) roots exist, so it is possible.

Solving: $x=\dfrac{-b}{2a}=\dfrac{40}{2}=20$x=-b/2a=40/2=20.

So the length is 20 m, and the breadth is $40-20=20$40-20=20 m — the park is a square.

Length = 20 m, Breadth = 20 m.