Solutions-Part-1-Class-10-Mathematics-1-Chapter-2-Quadratic Equations-Maharashtra Board

(Here students has to write any two equations in the form ax² + bx + c = 0, a ≠  0)

(i) x² − 8x + 5 = 0  (ii) 3y² = 27.

In given equation, x is the only variable with highest index 2.

∴ the given equation is a quadratic equation

In given equation, y is the only variable with highest index 2.

∴ the given equation is a quadratic equation

Multiplying the equation by y,

y³ + 1 = 2y  

∴ y³ − 2y + 1 = 0

In this equation, the highest index of variable y is 3 and not 2.

∴ the given equation is not a quadratic equation.

Multiplying the equation by x,

x² + 1= −2x 

∴ x² + 2x + 1 = 0.

In this equation, x is the only variable with highest index 2.

∴ the given equation is a quadratic equation.

(m + 2) (m− 5) = 0

 m(m−5) + 2(m−5) = 0

 m²−5m + 2m − 10 = 0

 m²−3m−10 = 0

In this equation, m is the only variable with highest index 2.

∴  the given equation is a quadratic equation.

m³ + 3 m² − 2 = 3m³

 m³ −3m³ + 3m² − 2 = 0

 −2m³ + 3m² − 2 = 0

In this equation, the highest index of the variable m is 3 and not 2.

∴ the given equation is not a quadratic equation.

2y = 10 − y²

y² + 2y − 10 = 0

Comparing with the standard form ax² + bx + c = 0

a = 1, b = 2, c = 10.

∴ The required equation is y² + 2y − 10 = 0,

Values of a, b, c is, a = 1, b = 2, c = 10.

(x − 1)² = 2 x + 3

∴ x² − 2x + 1−2x−3 = 0

∴ x² − 4x−2 = 0

Comparing with the standard form ax² + bx + c = 0

a = 1, b = −4, c = −2.

∴ The required equation is x² − 4x−2 = 0

Values of a, b, c is,  a = 1, b = −4, c = −2.

x² + 5x = −(3 − x)

 x² + 5x = −3 + x 

∴ x² + 5x − x + 3 = 0

 x² + 4x + 3 = 0

Comparing with the standard form ax² + bx + c = 0

a = 1, b = 4, c = 3.

∴ The required equation is  x² + 4x + 3 = 0

Values of a, b, c is,  a = 1, b = 4, c = 3.

3m² = 2 m² − 9

 3m²−2m² + 9 = 0

∴ m² + 0m + 9 = 0

Comparing with the standard form ax² + bx + c = 0

a =1, b = 0, c = 9.

∴ The required equation is m² + 0m + 9 = 0

Values of a, b, c is,  a = 1, b = 0, c = 9.

p(3 + 6p) = − 5

 3p + 6p² + 5 = 0

 6p² + 3p + 5 = 0

Comparing with the standard form ax² + bx + c = 0

a = 6, b = 3, c = 5.

∴ The required equation is 6p² + 3p + 5 = 0

Values of a, b, c is,  a = 6, b = 3, c = 5.

x² − 9 = 13

 x² − 9 − 13 = 0

 x² − 22 = 0

 ∴ x² + 0x −22 = 0

Comparing with the standard form ax² + bx + c = 0

a =1, b = 0, c = −22.

∴ The required equation is x² + 0x −22 = 0

Values of a, b, c is,  a =1, b = 0, c = −22.

Substituting x = 1,

LHS = (1)² + 4(1) − 5

         = 1 + 4 − 5

         = 5 − 5 = 0 = RHS

∴ x = 1 is the root of the given quadratic equation.

Substituting x = − 1,

LHS = (−1)² + 4(−1)−5

         = 1 − 4 − 5

          = 1 − 9 = − 8 ≠ RHS

∴ x = − 1 is not the root of the given quadratic equation.

∴  1 is the root; −1 is not the root.

Substituting m = 2,

LHS = 2(2)² − 5(2)

        = 2(4) − 10

         = 8 − 10

         = − 2 ≠ RHS

 ∴ m = 2 is not the root of the given quadratic equation.

Substituting m = \(\frac{5}{2}\),

LHS = \(2(\frac{5}{2})^2\)  − \(5(\frac{5}{2})\)

         = \(\frac{25}{2}-\frac{25}{2}\) = 0 = RHS

∴ m = \(\frac{5}{2}\) is the root of the given quadratic equation

∴ 2 is not the root; \(\frac{5}{2}\),  is the root.

x = 3 is the root of the given equation

Substituting x = 3 in the given equation,

k(3)² − 10(3) + 3 = 0

∴ 9k − 30 + 3 = 0

∴ 9k − 27 = 0

∴9k = 27

∴ k = 27/9 = 3

Answer : The value of k is 3

[\(-\frac{7}{5}\)] is a root of quadratic equation 5m² + 2m + k = 0

∴ Put m = [\(-\frac{7}{5}\)] in the equation.

5 × \([-\frac{7}{5}]^2\)  + 2 × [\(-\frac{7}{5}\)] + k = 0

[\(-\frac{49}{25}\) ]  + [\(-\frac{14}{5}\) ] + k = 0

[7] + k = 0

∴ k = [−7]

(In the question, boxes [−−] were to be fi1led.)

x² − 15 x + 54 = 0

∴ x² − 9x − 6x + 54 = 0         

∴ x(x−9)− 6(x−9) = 0

∴ (x−9)(x−6)=0

∴ x−9 = 0  or  x−6 = 0

∴ x = 9 or x = 6

Answer is : 9, 6 are the roots of the given quadratic equation.

x²  + x − 20 = 0

∴ x² + 5x − 4x − 20 = 0

∴ x(x + 5) − 4(x + 5) = 0

∴ (x + 5)(x − 4) = 0

 ∴ x + 5 = 0 or  x − 4 = 0

∴ x = −5 or x = 4

Answer is : −5, 4 are the roots of the given quadratic equation.

2y² + 27y + 13 = 0              …(..a × c = 2 × 13 = 26 …26 = 26 × 1 …b = 27 = 26 + 1)

∴ 2y² + 26y + y + 13 = 0 

∴ 2y(y + 13) + 1(y+13) = 0

∴  (y + 13)(2y + 1) = 0

∴ y + 13 = 0 or 2y + 1 = 0

∴ y = −13 or 2y = −1 ∴ y = \(-\frac{1}{2}\)

Answer is : − 13, \(-\frac{1}{2}\)  are the roots of the given quadratic equation

5m² = 22 m + 15                       

∴ 5m²−22m−15 = 0                    (..a x c = 5 × (−15) = −75 = −25 × 3, −25 + 3 = 22 =b)

∴.5m²−25m + 3m −l5 = 0

∴ 5m(m−5) + 3(m−5) = 0

∴ (m−5)(5m + 3) = 0 _

∴ m−5 = 0 or 5m + 3 = 0

∴ m = 5 or 5m = −3  ∴ m = \(-\frac{3}{5}\)

Answer is :  5, \(-\frac{3}{5}\)  are the roots of the given quadratic equation.

2x² − 2 x + \(\frac{1}{2}\)  = 0

∴ 2x² −x −x + \(\frac{1}{2}\) = 0                      ….( a × c = 2 × \(\frac{1}{2}\) = 1 = −1 × −1 … b = −1 −1 = −2)

∴ x(2x−1) − \(\frac{1}{2}\)(2x−1) = 0

 ∴  (2x−1) (x−\(\frac{1}{2}\) ) = 0

∴ 2x−1 = 0 or x − \(\frac{1}{2}\) = 0

∴ 2x = 1 or x = \(\frac{1}{2}\)

∴ x = \(\frac{1}{2}\) or x = \(\frac{1}{2}\)

Answer is : \(\frac{1}{2}\), \(\frac{1}{2}\) are the roots of the given quadratic equation.

[Note : Here, both the roots are the same.]

6x − \(\frac{1}{2}\) = 1

Multiplying each term of the equation by x,

6x² − 2 = x

6x²− x − 2 = 0

∴ 6x² − 4x + 3x − 2 = 0               ….( a × c =  6 × (−2) = −12 = −4 × 3 ..b = −4 + 3 = −1)

∴ 2x(3x−2) + 1(3x−2) = 0

∴ (3x−2)(2x+ 1) = 0

 ∴3x−2 = 0 or 2x + 1 = 0

∴3x = 2 or 2x = −1

∴ x = \(\frac{2}{3}\) or x = \(-\frac{1}{2}\)

Answer is: \(\frac{2}{3}\), \(-\frac{1}{2}\)  are the roots of the given quadratic equation.

\(\sqrt{2}\)x²  + 7 x + 5\(\sqrt{2}\)  = 0

\(\sqrt{2}\)x² + [5x] + [2x] + 5\(\sqrt{2}\)  = 0

x( \(\sqrt{2}\)x + 5) + \(\sqrt{2}\)(\(\sqrt{2}\)x + 5) = 0

( \(\sqrt{2}\)x + 5)(x + \(\sqrt{2}\) ) = 0

(\(\sqrt{2}\)x + 5) = 0 or (x + \(\sqrt{2}\) ) = 0

∴ x = \(-\frac{5}{\sqrt{2}}\) or x = \(-\sqrt{2}\)

∴ \(-\frac{5}{\sqrt{2}}\) and \(-\sqrt{2}\)  are roots of the equation.

3x² − 2\(\sqrt{6}\)x + 2 = 0

∴ 3x² − \(\sqrt{6}\)x − \(\sqrt{6}\)x + 2 = 0   (a × c =  3 × 2 = 6 = −   ..b =  )

∴ 3x² − \(\sqrt{3}\)x ×\(\sqrt{2}\) − \(\sqrt{3}\)x × \(\sqrt{2}\) + \(\sqrt{2}\) × \(\sqrt{2}\)  = 0   

∴ \(\sqrt{3}\)x(\(\sqrt{3}\)x − \(\sqrt{2}\)) − \(\sqrt{2}\)(\(\sqrt{3}\)x − \(\sqrt{2}\)) = 0

∴ (\(\sqrt{3}\)x − \(\sqrt{2}\))(\(\sqrt{3}\)x − \(\sqrt{2}\)) = 0

∴ \(\sqrt{3}\)x − \(\sqrt{2}\) = 0 or \(\sqrt{3}\)x − \(\sqrt{2}\) =0

∴ x = \(\frac{\sqrt{2}}{\sqrt{3}}\)

Here, both the roots are the same.

Answer is : \(\frac{\sqrt{2}}{\sqrt{3}}\)   are the roots of the given quadratic equation.

2m(m − 24) = 50.

∴ 2m² −48m − 50 = 0

∴ m² − 24m − 25 = 0      …..(Dividing each term of the equation by 2)

∴ m² − 25m + m − 25 = 0

∴ m(m − 25) + 1(m − 25) = 0

 ∴  (m − 25)(m + 1) = 0

∴ m − 25 = 0 or m + 1 = 0

∴ m = 25 or m = −1

Answer is : 25, − 1 are the roots of the given quadratic equation.

25m² = 9

∴ 25m² − 9 = 0

∴ (5m)² − (3)² = 0

∴ (5m + 3)(5m − 3) = 0

∴ 5m + 3 = 0 or 5m − 3 = 0

∴ 5m = −3 or 5m = 3

∴ m = \(-\frac{3}{5}\) or m = \(\frac{3}{5}\)

Answer is : \(-\frac{3}{5}\), \(\frac{3}{5}\) are the roots of the given quadratic equation.

7m² = 21m

∴ 7m² − 21m = 0

∴ 7m(m − 3) = 0

∴ 7m = 0 or m−3 = 0

∴ m = 0 or m = 3

Answer is : 0, 3 are the roots of the given quadratic equation.

m² − 11 = 0

∴ m² − \((\sqrt{11})^2\)  = 0

∴ (m + \(\sqrt{11}\) ) (m − \(\sqrt{11}\) ) = 0

∴ m + \(\sqrt{11}\) = 0 or m − \(\sqrt{11}\)  = 0

∴ m = − \(\sqrt{11}\)  or m =   \(\sqrt{11}\) 

Answer is : \(-\sqrt{11}\), \(\sqrt{11}\)  are the roots of the given quadratic equation.

x² + x − 20 = 0

Considering the first two terms on LHS, let’s find the third suitable square term to make the polynomial a perfect square.

Comparing x² + x with x² + 2xy,

2xy = x  ∴ 2y = 1 ∴ y = \(\frac{1}{2}\)  ∴ y² = \(\frac{1}{4}\)

∴ x² + x + \(\frac{1}{4}\) is a perfect square trinomial

x² + x−20 = 0

∴ x² + x + \(\frac{1}{4}\) − \(\frac{1}{4}\) − 20 = 0

∴ \((x +\frac{1}{2})^2-(\frac{1}{4}+20)\) = 0

∴ \((x +\frac{1}{2})^2-(\frac{1+80}{4})\) = 0

∴ \((x +\frac{1}{2})^2-(\frac{81}{4})\) = 0

∴ \((x +\frac{1}{2})^2-(\frac{9}{2})^2\) = 0

∴ \((x +\frac{1}{2}+\frac{9}{2})(x +\frac{1}{2}-\frac{9}{2})\) = 0

∴ \((x +\frac{1+9}{2})(x +\frac{1-9}{2})\) = 0

∴ \((x +\frac{10}{2})(x +\frac{-8}{2})\) = 0

∴ ( x + 5)(x − 4) = 0

∴ x + 5 = 0 or x − 4 = 0

∴ x = − 5 or x = 4

Answer is : −5, 4 are the roots of the given quadratic equation.

x² + 2x − 5 = 0

Considering the first two terms on LHS, let’s find the third suitable square term to make the polynomial a perfect square.

Comparing x² + 2x with x² + 2xy,

2xy = 2x  ∴ y = 1 ∴ y² = 1

 x² + 2x + 1 is a perfect square trinomial.

x² + 2x − 5 = 0

∴ x² + 2x + 1 − 1−5 = 0

∴ (x + 1)² − 6 = 0

∴ (x + 1)²−( )² = 0

∴(x + 1 + \(\sqrt{6}\))(x + 1− \(\sqrt{6}\)) = 0

∴ x + 1 + \(\sqrt{6}\) = 0 or x + 1− \(\sqrt{6}\) = 0

∴x = −1− \(\sqrt{6}\)  or x = −1 + \(\sqrt{6}\)

Answer is : −1−\(\sqrt{6}\) , −1 + \(\sqrt{6}\)   are the roots of the given quadratic equation.

m² − 5m = − 3

 m² − 5m + 3 = 0            ….(Standard form)

Considering the first two terms on LHS, let’s find the third suitable square term to make the polynomial a perfect square.

Comparing m² − 5m with m² − 2mn,

− 2mn = − 5m

∴ n = \(\frac{5}{2}\) ∴ n² = \(\frac{25}{4}\)

 m² − 5m + \(\frac{25}{4}\) is a perfect square trinomial.

m² − 5m + 3 = 0

m² − 5m + \(\frac{25}{4}-\frac{25}{4}\) + 3 = 0

∴ \((m-\frac{5}{2})^2-(\frac{25}{4}-3)\) = 0

∴ \((m-\frac{5}{2})^2-(\frac{25-12}{4})\) = 0

∴ \((m-\frac{5}{2})^2-(\frac{13}{4})\) = 0

∴ \((m-\frac{5}{2})^2-(\frac{\sqrt{13}}{2})^2\) = 0

∴ \((m-\frac{5}{2} +\frac{\sqrt{13}}{2})(m-\frac{5}{2} -\frac{\sqrt{13}}{2})\) = 0

∴ \((m-\frac{5}{2} +\frac{\sqrt{13}}{2})\) = 0 or \((m-\frac{5}{2} -\frac{\sqrt{13}}{2})\)  = 0

∴ m = \(\frac{5}{2} -\frac{\sqrt{13}}{2}\)  or m = \(\frac{5}{2} +\frac{\sqrt{13}}{2}\)

∴ m = \(\frac{5-\sqrt{13}}{2}\)  or m = \(\frac{5+\sqrt{13}}{2}\)

Answer is : \(\frac{5-\sqrt{13}}{2}\), \(\frac{5+\sqrt{13}}{2}\)  are the roots of the given quadratic equation.

9y² − 12 y + 2 = 0

It is convenient to make the coefficient of quadratic term 1

∴ y² − \(\frac{12y}{9} +\frac{2}{9}\) = 0                 …. (Dividing by 9)

∴ y² − \(\frac{4y}{3} +\frac{2}{9}\) = 0                 

Comparing y² − \(\frac{4y}{3}\) with y² − 2yz,

−2yz = − \(\frac{4y}{3}\)   ∴ z = \(\frac{2}{3}\)  ∴ z² = \(\frac{4}{9}\)

 ∴ y² − \(\frac{4y}{3}\) + \(\frac{4}{9}\) is a perfect square trinomial.

y² − \(\frac{4y}{3} +\frac{2}{9}\) = 0                 

∴ y² − \(\frac{4y}{3} +\frac{4}{9}-\frac{4}{9}+\frac{2}{9}\)  = 0                 

∴ \((y-\frac{2}{3})^2-(\frac{4}{9}-\frac{2}{9}\) = 0

∴ \((y-\frac{2}{3})^2-(\frac{4-2}{9})\) = 0

∴ \((y-\frac{2}{3})^2-(\frac{2}{9})\) = 0

∴ \((y-\frac{2}{3})^2-(\frac{\sqrt{2}}{3})^2\) = 0

∴ \((y-\frac{2}{3}+\frac{\sqrt{2}}{3})(y-\frac{2}{3}-\frac{\sqrt{2}}{3})\) = 0

∴ \((y-\frac{2}{3}+\frac{\sqrt{2}}{3})\) = 0 or \((y-\frac{2}{3}-\frac{\sqrt{2}}{3})\) = 0

∴ y = \(\frac{2}{3}-\frac{\sqrt{2}}{3}\)  or y = \(\frac{2}{3}+\frac{\sqrt{2}}{3}\)

∴ y = \(\frac{2-\sqrt{2}}{3}\)  or y = \(\frac{2+\sqrt{2}}{3}\)

Answer is : \(\frac{2-\sqrt{2}}{3}\), \(\frac{2+\sqrt{2}}{3}\) are the roots of the given quadratic equation.

2y² + 9y + 10 = 0

∴ y² + \(\frac{9}{2}\)y + 5 = 0        ……(Dividing by 2)

Comparing y² + \(\frac{9}{2}\)y with y² + 2yz,

2yz = \(\frac{9}{2}\)y  ∴ z = \(\frac{9}{4}\), z² = \(\frac{81}{16}\)

∴ y² + \(\frac{9}{2}\)y + \(\frac{81}{16}\) is a perfect square trinomial.

y² + \(\frac{9}{2}\)y + 5 = 0  

∴ y² + \(\frac{9}{2}\)y + \(\frac{81}{16}\) - \(\frac{81}{16}\) + 5 = 0        

∴ \((y+\frac{9}{4})^2-(\frac{81}{16}+5)\) = 0

∴ \((y+\frac{9}{4})^2-(\frac{81-80}{16})\) = 0

∴ \((y+\frac{9}{4})^2-(\frac{1}{16})\) = 0

∴ \((y+\frac{9}{4})^2-(\frac{1}{4})^2\) = 0

∴ \((y+\frac{9}{4}+\frac{1}{4})(y+\frac{9}{4}-\frac{1}{4})\) = 0

∴ \((y+\frac{9+1}{4})(y+\frac{9-1}{4})\) = 0

∴ \((y+\frac{10}{4})(y+\frac{8}{4})\) = 0

∴ \(y+\frac{10}{4}\) = 0 or \(y+\frac{8}{4}\) = 0

∴ y = \(-\frac{10}{4}\) = 0 or y= \(-\frac{8}{4}\) 

∴ y = \(-\frac{5}{2}\) or y = −2

Answer is : \(-\frac{5}{2}\), −2 are the roots of the given quadratic equation.

5x² = 4x + 7

∴ 5x² − 4x − 7 = 0

x² − \(\frac{4}{5}\)x − \(\frac{7}{5}\) = 0     …. (Dividing by 5)

Comparing x² − \(\frac{4}{5}\)x with x² − 2xy,

− 2xy = − \(\frac{4}{5}\)x  ∴ y = \(\frac{2}{5}\), ∴ y² = \(\frac{4}{25}\)

x² − \(\frac{4}{5}\)x + \(\frac{4}{25}\)  is a perfect square trinomial.

x² − \(\frac{4}{5}\)x − \(\frac{7}{5}\)

∴ x² − \(\frac{4}{5}\)x + \(\frac{4}{25}\) - \(\frac{4}{25}\)  − \(\frac{7}{5}\) = 0    

∴ \((x-\frac{2}{5})^2-(\frac{4}{25}+\frac{7}{5})\) = 0

∴ \((x-\frac{2}{5})^2-(\frac{4+35}{25})\) = 0

∴ \((x-\frac{2}{5})^2-(\frac{39}{25})\) = 0

∴ \((x-\frac{2}{5})^2-(\frac{\sqrt{39}}{5})^2\) = 0

∴ \((x-\frac{2}{5}+\frac{\sqrt{39}}{5})(x-\frac{2}{5}-\frac{\sqrt{39}}{5})\)  = 0

∴ \(x-\frac{2}{5}+\frac{\sqrt{39}}{5}\) = 0 or \(x-\frac{2}{5}-\frac{\sqrt{39}}{5}\) = 0

∴ x = \(\frac{2}{5}-\frac{\sqrt{39}}{5}\) or x = \(\frac{2}{5}+\frac{\sqrt{39}}{5}\)

∴ x = \(\frac{2-\sqrt{39}}{5}\) or x = \(\frac{2+\sqrt{39}}{5}\)

Answer is : \(\frac{2-\sqrt{39}}{5}\), \(\frac{2+\sqrt{39}}{5}\) are the roots of the given quadratic equation.