Quadratic Equations
Class10Mathematics1Chapter2Maharashtra Board
SolutionsPart1
Solutions Part1

Practice Set 2.1
Question 1. Write any two quadratic equations.
(Here students has to write any two equations in the form ax^{2} + bx + c = 0, a ≠ 0)
(i) x^{2} − 8x + 5 = 0 (ii) 3y^{2} = 27.
Question 2. Decide which of the following are quadratic equations.
(1) x^{2} + 5 x − 2 = 0
In given equation, x is the only variable with highest index 2.
∴ the given equation is a quadratic equation
(2) y^{2} = 5 y − 10
In given equation, y is the only variable with highest index 2.
∴ the given equation is a quadratic equation
(3) y^{2} + \(\frac{1}{y}\) = 2
Multiplying the equation by y,
y^{3} + 1 = 2y
∴ y^{3} − 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.
(4) x + \(\frac{1}{x}\) = −2
Multiplying the equation by x,
x^{2} + 1= −2x
∴ x^{2} + 2x + 1 = 0.
In this equation, x is the only variable with highest index 2.
∴ the given equation is a quadratic equation.
(5) (m + 2) (m − 5) = 0
(m + 2) (m− 5) = 0
m(m−5) + 2(m−5) = 0
m^{2}−5m + 2m − 10 = 0
m^{2}−3m−10 = 0
In this equation, m is the only variable with highest index 2.
∴ the given equation is a quadratic equation.
(6) m^{3} + 3 m^{2} − 2 = 3m^{3}
m^{3} + 3 m^{2} − 2 = 3m^{3}
m^{3} −3m^{3} + 3m^{2} − 2 = 0
−2m^{3} + 3m^{2} − 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.
Question 3. Write the following equations in the form ax^{2} + bx + c = 0, then write the values of a, b, c for each equation.
(1) 2y = 10 − y^{2}
2y = 10 − y^{2}
y^{2} + 2y − 10 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a = 1, b = 2, c = 10.
∴ The required equation is y^{2} + 2y − 10 = 0,
Values of a, b, c is, a = 1, b = 2, c = 10.
(2) (x − 1)^{2} = 2 x + 3
(x − 1)^{2} = 2 x + 3
∴ x^{2} − 2x + 1−2x−3 = 0
∴ x^{2} − 4x−2 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a = 1, b = −4, c = −2.
∴ The required equation is x^{2} − 4x−2 = 0
Values of a, b, c is, a = 1, b = −4, c = −2.
(3) x^{2} + 5x = −(3 − x)
x^{2} + 5x = −(3 − x)
x^{2} + 5x = −3 + x
∴ x^{2} + 5x − x + 3 = 0
x^{2} + 4x + 3 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a = 1, b = 4, c = 3.
∴ The required equation is x^{2} + 4x + 3 = 0
Values of a, b, c is, a = 1, b = 4, c = 3.
(4) 3m^{2} = 2 m^{2} − 9
3m^{2} = 2 m^{2} − 9
3m^{2}−2m^{2} + 9 = 0
∴ m^{2} + 0m + 9 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a =1, b = 0, c = 9.
∴ The required equation is m^{2} + 0m + 9 = 0
Values of a, b, c is, a = 1, b = 0, c = 9.
(5) P (3 + 6p) = −5
p(3 + 6p) = − 5
3p + 6p^{2} + 5 = 0
6p^{2} + 3p + 5 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a = 6, b = 3, c = 5.
∴ The required equation is 6p^{2} + 3p + 5 = 0
Values of a, b, c is, a = 6, b = 3, c = 5.
(6) x^{2} − 9 = 13
x^{2} − 9 = 13
x^{2 }− 9 − 13 = 0
x^{2} − 22 = 0
∴ x^{2} + 0x −22 = 0
Comparing with the standard form ax^{2} + bx + c = 0
a =1, b = 0, c = −22.
∴ The required equation is x^{2} + 0x −22 = 0
Values of a, b, c is, a =1, b = 0, c = −22.
Question 4. Determine whether the values given against each of the quadratic equation are the roots of the equation.
(1) x^{2} + 4x − 5 = 0 , x = 1, −1
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.
(2) 2m^{2} − 5m = 0 , m = 2, \(\frac{5}{2}\)
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.
Question 5. Find k if x = 3 is a root of equation kx^{2} − 10x + 3 = 0 .
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
Question 6. One of the roots of equation 5m^{2} + 2m + k = 0 is \(\frac{7}{5}\), . Complete the following activity to find the value of ’k’.
[\(\frac{7}{5}\)] is a root of quadratic equation 5m^{2} + 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.)
Practice Set 2.2
Question. Solve the following quadratic equations by factorisation.
(1) x^{2} − 15 x + 54 = 0
x^{2} − 15 x + 54 = 0
∴ x^{2} − 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.
(2) x^{2 }+ x − 20 = 0
x^{2 }+ x − 20 = 0
∴ x^{2} + 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.
(3) 2y^{2} + 27y + 13 = 0
2y^{2} + 27y + 13 = 0 …(..a × c = 2 × 13 = 26 …26 = 26 × 1 …b = 27 = 26 + 1)
∴ 2y^{2} + 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
(4) 5m^{2 }= 22 m + 15
5m^{2 }= 22 m + 15
∴ 5m^{2}−22m−15 = 0 (..a x c = 5 × (−15) = −75 = −25 × 3, −25 + 3 = 22 =b)
∴.5m^{2}−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.
(5) 2x^{2} − 2 x + \(\frac{1}{2}\) = 0
2x^{2} − 2 x + \(\frac{1}{2}\) = 0
∴ 2x^{2 }−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.]
(6) 6x − \(\frac{2}{x}\) = 1
6x − \(\frac{1}{2}\) = 1
Multiplying each term of the equation by x,
6x^{2} − 2 = x
6x^{2}− x − 2 = 0
∴ 6x^{2} − 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.
(7) \(\sqrt{2}\)x^{2 } + 7 x + 5\(\sqrt{2}\) = 0 to solve this quadratic equation by factorisation, complete the following activity.
\(\sqrt{2}\)x^{2 } + 7 x + 5\(\sqrt{2}\) = 0
\(\sqrt{2}\)x^{2} + [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.
(8) 3x^{2} − 2\(\sqrt{6}\)x + 2 = 0
3x^{2} − 2\(\sqrt{6}\)x + 2 = 0
∴ 3x^{2} − \(\sqrt{6}\)x − \(\sqrt{6}\)x + 2 = 0 (a × c = 3 × 2 = 6 = − ..b = )
∴ 3x^{2} − \(\sqrt{3}\)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.
(9) 2m (m − 24) = 50
2m(m − 24) = 50.
∴ 2m^{2} −48m − 50 = 0
∴ m^{2 }− 24m − 25 = 0 …..(Dividing each term of the equation by 2)
∴ m^{2} − 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.
(10) 25m^{2} = 9
25m^{2} = 9
∴ 25m^{2} − 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.
(11) 7m^{2} = 21m
7m^{2} = 21m
∴ 7m^{2} − 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.
(12) m^{2} − 11 = 0
m^{2} − 11 = 0
∴ m^{2} − \((\sqrt{11})^2\) = 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.
Practice Set 2.3
Solve the following quadratic equations by completing the square method.
(1) x^{2} + x − 20 = 0
x^{2} + 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 x^{2} + x with x^{2} + 2xy,
2xy = x ∴ 2y = 1 ∴ y = \(\frac{1}{2}\) ∴ y^{2} = \(\frac{1}{4}\)
∴ x^{2} + x + \(\frac{1}{4}\) is a perfect square trinomial
x^{2} + x−20 = 0
∴ x^{2} + 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
∴ \((x +\frac{1+9}{2})(x +\frac{19}{2})\) = 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.
(2) x^{2} + 2x − 5 = 0
x^{2} + 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 x^{2} + 2x with x^{2} + 2xy,
2xy = 2x ∴ y = 1 ∴ y^{2} = 1
x^{2} + 2x + 1 is a perfect square trinomial.
x^{2} + 2x − 5 = 0
∴ x^{2} + 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.
(3) m^{2} − 5m = −3
m^{2} − 5m = − 3
m^{2} − 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 m^{2} − 5m with m^{2} − 2mn,
− 2mn = − 5m
∴ n = \(\frac{5}{2}\) ∴ n^{2} = \(\frac{25}{4}\)
m^{2} − 5m + \(\frac{25}{4}\) is a perfect square trinomial.
m^{2} − 5m + 3 = 0
m^{2} − 5m + \(\frac{25}{4}\frac{25}{4}\) + 3 = 0
∴ \((m\frac{5}{2})^2(\frac{25}{4}3)\) = 0
∴ \((m\frac{5}{2})^2(\frac{2512}{4})\) = 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.
(4) 9y^{2} − 12 y + 2 = 0
9y^{2} − 12 y + 2 = 0
It is convenient to make the coefficient of quadratic term 1
∴ y^{2} − \(\frac{12y}{9} +\frac{2}{9}\) = 0 …. (Dividing by 9)
∴ y^{2} − \(\frac{4y}{3} +\frac{2}{9}\) = 0
Comparing y^{2} − \(\frac{4y}{3}\) with y^{2} − 2yz,
−2yz = − \(\frac{4y}{3}\) ∴ z = \(\frac{2}{3}\) ∴ z^{2} = \(\frac{4}{9}\)
∴ y^{2} − \(\frac{4y}{3}\) + \(\frac{4}{9}\) is a perfect square trinomial.
y^{2} − \(\frac{4y}{3} +\frac{2}{9}\) = 0
∴ y^{2} − \(\frac{4y}{3} +\frac{4}{9}\frac{4}{9}+\frac{2}{9}\) = 0
∴ \((y\frac{2}{3})^2(\frac{4}{9}\frac{2}{9}\) = 0
∴ \((y\frac{2}{3})^2(\frac{42}{9})\) = 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.
(5) 2y^{2} + 9y + 10 = 0
2y^{2} + 9y + 10 = 0
∴ y^{2} + \(\frac{9}{2}\)y + 5 = 0 ……(Dividing by 2)
Comparing y^{2} + \(\frac{9}{2}\)y with y^{2} + 2yz,
2yz = \(\frac{9}{2}\)y ∴ z = \(\frac{9}{4}\), z^{2} = \(\frac{81}{16}\)
∴ y^{2} + \(\frac{9}{2}\)y + \(\frac{81}{16}\) is a perfect square trinomial.
y^{2} + \(\frac{9}{2}\)y + 5 = 0
∴ y^{2} + \(\frac{9}{2}\)y + \(\frac{81}{16}\)  \(\frac{81}{16}\) + 5 = 0
∴ \((y+\frac{9}{4})^2(\frac{81}{16}+5)\) = 0
∴ \((y+\frac{9}{4})^2(\frac{8180}{16})\) = 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
∴ \((y+\frac{9+1}{4})(y+\frac{91}{4})\) = 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.
(6) 5x^{2} = 4x + 7
5x^{2} = 4x + 7
∴ 5x^{2} − 4x − 7 = 0
x^{2} − \(\frac{4}{5}\)x − \(\frac{7}{5}\) = 0 …. (Dividing by 5)
Comparing x^{2} − \(\frac{4}{5}\)x with x^{2} − 2xy,
− 2xy = − \(\frac{4}{5}\)x ∴ y = \(\frac{2}{5}\), ∴ y^{2} = \(\frac{4}{25}\)
x^{2} − \(\frac{4}{5}\)x + \(\frac{4}{25}\) is a perfect square trinomial.
x^{2} − \(\frac{4}{5}\)x − \(\frac{7}{5}\)
∴ x^{2} − \(\frac{4}{5}\)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.
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