Class 10: Quadratic Equations – Miscellaneous Problems – Set 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Solve:

(i) $2x^2 - 7x = 39$ (ii) $x^2 = 5x$ (iii) $x^2 = 16$

Answer:

(i) $\displaystyle 2x^2 - 7x = 39$

$\displaystyle \Rightarrow 2x^2 - 7x - 39 = 0$

$\displaystyle \Rightarrow 2x^2 - 13 x + 6 x - 39 = 0$

$\displaystyle \Rightarrow x(2x-13) + 3( 2x - 13) = 0$

$\displaystyle \Rightarrow (2x-13)(x+3) = 0$

$\displaystyle \Rightarrow 2x-13 = 0 \text{ or } x + 3 = 0$

$\displaystyle \Rightarrow x = \frac{13}{2} \text{ or } x = - 3$

(ii) $\displaystyle x^2 = 5x$

$\displaystyle \Rightarrow x^2 - 5x = 0$

$\displaystyle \Rightarrow x(x-5) = 0$

$\displaystyle \Rightarrow x = 0 \text{ or } x -5 = 0$

$\displaystyle \Rightarrow x = 0 \text{ or } x = 5$

(iii) $\displaystyle x^2 = 16$

$\displaystyle \Rightarrow x^2 - 16 = 0$

$\displaystyle \Rightarrow (x+4)(x-4) = 0$

$\displaystyle \Rightarrow x+ 4 = 0 \text{ or } x - 4 = 0$

$\displaystyle \Rightarrow x = -4 \text{ or } x = 4$

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$\displaystyle \text{Question 2: Solve: } \frac{x}{x-1} + \frac{x-1}{x} = 2\frac{1}{2}$

Answer:

$\displaystyle \frac{x}{x-1} + \frac{x-1}{x} = 2\frac{1}{2}$

$\displaystyle \Rightarrow \frac{x^2 + ( x-1)^2}{x(x-1)} = \frac{5}{2}$

$\displaystyle \Rightarrow 2 ( x^2 + x^2 - 2x + 1) = 5 ( x^2 - x)$

$\displaystyle \Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x$

$\displaystyle \Rightarrow -x^2 + x + 2 = 0$

$\displaystyle \Rightarrow x^2 - x - 2 = 0$

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

$\displaystyle \Rightarrow x-2 = 0 \text{ or } x+1 = 0$

$\displaystyle \Rightarrow x = 2 \text{ or } x = -1$

$\\$

Question 3: Find the quadratic equation whose solution set is {-2, 3}.

Answer:

Since, solution set $\displaystyle = \{ -2, 3 \}$

$\displaystyle \Rightarrow x=-2, \text{ or } x=3$

$\displaystyle \Rightarrow x+2=0, \text{ or } x- 3 = 0$

$\displaystyle \Rightarrow (x+2) (x-3) =0$

$\displaystyle \Rightarrow x^2-3x+2x-6=0$

$\displaystyle \Rightarrow x^2 - x - 6 = 0$ which is the required quadratic equation.

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Question 4: Use the substitution $\displaystyle x = 3y + 1$ to solve for $y$ , if $5(3y+1)^2 + 6(3y+1) - 8 = 0$

Answer:

$\displaystyle 5(3y+1)^2 + 6(3y+1) - 8 = 0$

$\displaystyle \text{Putting } x = 3y+1$

$\displaystyle \Rightarrow 5x^2 + 6x - 9 = 0$

$\displaystyle \Rightarrow (x+2)(5x-4) = 0$

$\displaystyle \Rightarrow x = - 2 \text{ or } x = \frac{4}{5}$

$\displaystyle \Rightarrow \text{When } x = - 2, \text{ then } 3y+1 = - 2 \Rightarrow y = -1$

$\displaystyle \Rightarrow \text{ and when } x = \frac{4}{5}, \text{ then } 3y+1 = \frac{4}{5} \Rightarrow y = \frac{-1}{15}$

$\displaystyle \Rightarrow \text{ Therefore } y = -1 \text{ or } y = \frac{-1}{15}$

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Question 5: Without solving the quadratic equation $3x^2 - 2x- 1 = 0,$ find whether $x = 1$ is a solution (root) of this equation or not.

Answer:

Substituting $x = 1$ in the given equation $3x^2 - 2x - 1 = 0,$ we get :

$3(1)^2 - 2(1) -1 = 0$

$\Rightarrow 3 -2 -1 = 0$

Which is true.

Therefore $x = 1$ is a solution of the given equation $3x^2 - 2x- 1 = 0$

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Question 6: Without solving the quadratic equation $x^2 - x + 1 = 0,$ find whether $x = -1$ is a solution (root) of this equation or not.

Answer:

Substituting $x = -1$ in the given equation $x^2 - x + 1 = 0,$ we get :

$(-1)^2 -(-1)+1 = 0$

$\Rightarrow 1+1+1 = 0$

Which is not true.

Therefore $x =- 1$ is a solution of the given equation $x^2 - x + 1 = 0$

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Question 7: Find the value of $k$ for which $x = 2$ is a root (solution) of equation $kx^2 + 2x - 3 = 0.$

Answer:

Substituting $x = 2$ in the given equation $kx^2 + 2x - 3 = 0,$ we get :

$k(2)^2 + 2 \times 2 - 3 = 0$

$\Rightarrow 4k + 4 - 3 = 0$

$\Rightarrow 4k = -1$

$\displaystyle \Rightarrow k = \frac{-1}{4}$

$\\$

Question 8: If $x = 2$ and $x = 3$ are roots of the equation $3x^2 - 2mx + 2n = 0,$ find the values of $m$ and $n.$

Answer:

$x=2$ is a root of the equation $3x^2 - 2mx + 2n = 0$

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

$\Rightarrow 12 - 4m + 2n = 0$

$\Rightarrow - 4m + 2n =- 12$

$\Rightarrow 2m - n = 6 \text{ ... ... ... ... ... i)}$

$x=3$ is a root of the equation $3x^2 - 2mx + 2n = 0$

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

$\Rightarrow 27 - 6m + 2n = 0$

$\Rightarrow - 6m + 2n =- 27$

$\Rightarrow 6m - 2n = 27 \text{ ... ... ... ... ... ii)}$

On solving equations I and II, we get $m = 7.5$ and $n = 9$

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Question 9: If one root of the quadratic equation $2x^2 + ax - 6 = 0$ is 2 , find the value of $a.$ Also, find the other root.

Answer:

Since, $x=2$ is a root of the given equation $2x^2 + ax - 6 = 0$

$\Rightarrow 2(2)^2 + a \times 2 - 6 = 0$

$\Rightarrow 8 + 2a - 6 = 0$

$\Rightarrow a = - 1$

Substituting $a = -1,$ we get :

$2x^2 + ( -1) x - 6 = 0$

$\Rightarrow 2x^2 - x - 6 = 0$

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

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

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

$\Rightarrow x-2 = 0 \text{ or } 2x + 3 = 0$

$\displaystyle \Rightarrow x = 2 \text{ or } x = \frac{-3}{2}$

$\displaystyle \Rightarrow \text{The other root is } \frac{-3}{2}$

$\\$

Question 10: Solve each of the following equations by using the formula:

(i) $5x^2 - 2x - 3 = 0$ (ii) $x^2 = 18x - 77$ (iii) $\sqrt{3} x^2 + 11 x + 6\sqrt{3} =0$

Answer:

$\displaystyle \text{(i) Comparing } 5x^2 - 2x - 3 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 5, b = -2 \text{ and } c= -3$

$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values we get

$\displaystyle x = \frac{2 \pm \sqrt{(-2)^2 - 4(5)(-3)}}{2(5)} = \frac{2 \pm \sqrt{64}}{10} = \frac{2\pm8}{10}$

$\displaystyle \Rightarrow x = 1 \text{ and } x = \frac{-3}{5}$

$\\$

$\displaystyle \text{(ii) Comparing } x^2 = 18x - 77 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -18 \text{ and } c= 77$

$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values we get

$\displaystyle x = \frac{18 \pm \sqrt{(-18)^2 - 4(1)(77)}}{2(1)} = \frac{18 \pm \sqrt{16}}{2} = \frac{18\pm4}{2}$

$\displaystyle \Rightarrow x = 11 \text{ and } x = 7$

$\\$

$\displaystyle \text{(ii) Comparing } \sqrt{3} x^2 + 11 x + 6\sqrt{3} =0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = \sqrt{3}, b = 11 \text{ and } c= 6\sqrt{3}$

$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values we get

$\displaystyle x = \frac{-11 \pm \sqrt{(11)^2 - 4(\sqrt{3})(6\sqrt{3})}}{2(\sqrt{3})} = \frac{-11 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-11\pm 7}{2\sqrt{3}}$

$\displaystyle \Rightarrow x = \frac{-2\sqrt{3}}{3} \text{ and } x = -3\sqrt{3}$

$\\$

Question 11: Witt out solving, examine the nature of the roots of the equations :

(i) $5x^2 - 6x +7 = 0$ (ii) $x^2+6x+9=0$ (iii) $2x^2+6x+3=0$

Answer:

$\displaystyle \text{(i) Comparing } 5x^2 - 6x +7 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 5, b = -6 \text{ and } c= 7$

Therefore Discriminant $= b^2 - 4ac = (-6)^2 - 4 (5)(7) = 36 - 140 = - 104;$ which is negative.

$\displaystyle \text{Since } a, b, \text{ and } c \text{ are real numbers, } a \neq 0 \text{ and } b^2 - 4ac < 0$

Therefore the roots are not real i.e the roots are imaginary.

$\\$

$\displaystyle \text{(ii) Comparing } x^2+6x+9=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = 6 \text{ and } c= 9$

Therefore Discriminant $= b^2 - 4ac = (6)^2 - 4 (1)(9) = 36-36 = 0;$

$\displaystyle \text{Since } a, b, \text{ and } c \text{ are real numbers, } a \neq 0 \text{ and } b^2 - 4ac = 0$

Therefore the roots are equal.

$\\$

$\displaystyle \text{(iii) Comparing } 2x^2+6x+3=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 2, b = 6 \text{ and } c= 3$

Therefore Discriminant $= b^2 - 4ac = (6)^2 - 4 (2)(3) = 36-24 = 12;$ which is positive.

$\displaystyle \text{Since } a, b, \text{ and } c \text{ are real numbers, } a \neq 0 \text{ and } b^2 - 4ac > 0$

Therefore the roots are real and unequal.

$\\$

Question 12: Find the value of $'p'$ is the roots of the following quadratic equation are equal: $(p-3)x^2 + 6x + 9 = 0$

Answer:

$\displaystyle \text{Comparing } (p-3)x^2 + 6x + 9 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = (p-3), b = 6 \text{ and } c= 9$

Since the roots are equal, the discriminant is $0$

$\therefore b^2 - 4ac = 0$

$\Rightarrow (6)^2 - 4 ( p-3) (9) = 0$

$\Rightarrow 36 - 36 p + 108 = 0$

$\Rightarrow p = 4$

$\\$

Question 13: Find the value of $'m'$ is the roots of the following quadratic equation are equal: $(4+m)x^2 + ( m+1)x + 1 = 0$

Answer:

$\displaystyle \text{Comparing } (4+m)x^2 + ( m+1)x + 1 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = (4+m), b = (m+1) \text{ and } c= 1$

Since the roots are equal, the discriminant is $0$

$\therefore b^2 - 4ac = 0$

$\Rightarrow (m+1)^2 - 4 ( 4+m) (1) = 0$

$\Rightarrow m^2 + 2m + 1 - 16 - 4m = 0$

$\Rightarrow m^2 - 2m - 15 = 0$

$\Rightarrow (m-5)(m+3) = 0$

Therefore $m = 5$ or $m = - 3$

$\\$

Question 14: Solve each of the following equations for x and give, in each case, your answer correct to 2 decimal places.

(i) $x^2 - 10x + 6= 0$ (ii) $3x^2+5x-9=0$

Answer:

$\displaystyle \text{(i) Comparing } x^2 - 10x + 6= 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -10 \text{ and } c= 6$

$\therefore b^2 - 4ac = (-10)^2 - 4 (1)(6) = 100 - 24 = 76$

$\Rightarrow \sqrt{b^2-4ac} = \sqrt{76} = 8.718$

$\displaystyle \therefore x = \frac{10 \pm 8.718}{2(1)}$

$\displaystyle \Rightarrow x = 9.359 \text{ and } x = 0.641$

or correct to 2 decimal places: $\displaystyle x = 9.36 \text{ and } x = 0.64$

$\\$

$\displaystyle \text{(i) Comparing } 3x^2+5x-9=0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 3, b = 5 \text{ and } c= -9$

$\therefore b^2 - 4ac = (5)^2 - 4 (5)(-9) = 25+108 = 133$

$\Rightarrow \sqrt{b^2-4ac} = \sqrt{133} = 11.533$

$\displaystyle \therefore x = \frac{-5 \pm 11.533}{2(3)}$

$\displaystyle \Rightarrow x = 1.089 \text{ and } x =-2.756$

or correct to 2 decimal places: $\displaystyle x = 1.09 \text{ and } x = -2.76$

$\\$

$\displaystyle \text{Question 15: Solve the following equation: } x - \frac{18}{x} = 6.$

Give your answer correct to two significant figures. [ICSE Board 2011]

Answer:

$\displaystyle x - \frac{18}{x} = 6 \hspace{1.0cm} \Rightarrow x^2 - 18 = 6x \hspace{1.0cm} \Rightarrow x^2 - 6x - 18 = 0$

$\displaystyle \text{Comparing } x^2 - 6x - 18 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -6 \text{ and } c= -18$

$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting the values we get

$\displaystyle x = \frac{6 \pm \sqrt{(6)^2 - 4(1)(-18)}}{2(1)} = \frac{6 \pm 10.392}{2}$

$\displaystyle \Rightarrow x = 8.196 \text{ and } x = -2.196$

or correct to 2 decimal places: $\displaystyle x = 8.2 \text{ and } x = -2.2$

$\\$

Question 16: Solve:

(i) $2x^4- 5x^2 + 3 = 0$ (ii) $(x^2+3x)^2 - ( x^2 + 3x) - 6 = 0 , x \in R$

Answer:

(i) $2x^4- 5x^2 + 3 = 0$

Let $\displaystyle x^2 = y.$ Substituting:

$\displaystyle 2y^2 - 5y + 3 = 0$

$\displaystyle \Rightarrow (y-1)(2y-3) = 0$

$\displaystyle \Rightarrow y = 1 \text{ and } y = \frac{3}{2}$

$\displaystyle \text{When } y = 1 \text{ then } x^2 = 1 \hspace{1.0cm}\Rightarrow x = \pm 1$

$\displaystyle \text{When } y = \frac{3}{2} \text{ then } x^2 = \frac{3}{2} \hspace{1.0cm}\Rightarrow x = \pm \sqrt{ \frac{3}{2} }$

$\displaystyle \text{Therefore the required solution is } 1, -1, \sqrt{ \frac{3}{2} }, - \sqrt{ \frac{3}{2} }$

$\\$

(ii) $(x^2+3x)^2 - ( x^2 + 3x) - 6 = 0$

Let $\displaystyle x^2+3x = y.$ Substituting:

$\displaystyle y^2 - y -6 = 0$

$\displaystyle \Rightarrow (y-3)(2y+2) = 0$

$\displaystyle \Rightarrow y = 3 \text{ and } y = -2$

$\displaystyle \text{When } y = 3 \text{ then } x^2+3x = 3 \hspace{1.0cm}\Rightarrow x^2 + 3x - 3 =0$

$\displaystyle x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(-3)}}{2(1)} = \frac{-3 \pm \sqrt{21}}{2}$

$\displaystyle \text{When } y = -2 \text{ then } x^2+3x = -2 \hspace{1.0cm}\Rightarrow x^2 + 3x +2 =0$

$\displaystyle x = \frac{-3 \pm \sqrt{(3)^2 - 4(1)(2)}}{2(1)} = \frac{-3 \pm 1}{2} =-1 \text{ or } -2$

$\displaystyle \text{Therefore the required solution is } : \frac{-3 + \sqrt{21}}{2}, \frac{-3 - \sqrt{21}}{2} , -1 , -2$

$\\$

$\displaystyle \text{Question 17: Solve} : \sqrt{\frac{x}{1-x}} + \sqrt{\frac{1-x}{x}} = 2 \frac{1}{6}, x \neq 0 \text{ and } x \neq 1$

Answer:

$\displaystyle \text{Let } \sqrt{\frac{x}{1-x}} = y$

$\displaystyle \Rightarrow \sqrt{\frac{1-x}{x}} = \frac{1}{y}$

Therefore Given equation reduces to:

$\displaystyle y + \frac{1}{y} = \frac{13}{6}$

$\displaystyle \Rightarrow 6y^2 + 6 = 13y$

$\displaystyle \Rightarrow 6y^2 - 13y + 6 = 0$

$\displaystyle \Rightarrow ( 2y-3)(3y-2) = 0$

$\displaystyle \Rightarrow y = \frac{3}{2} , \text{ or } y = \frac{2}{3}$

$\displaystyle \text{When } y = \frac{3}{2} \hspace{1.0cm}\Rightarrow \sqrt{\frac{x}{1-x}} = \frac{3}{2} \hspace{1.0cm}\Rightarrow \frac{x}{1-x} = \frac{9}{4}$

$\displaystyle\Rightarrow 4x = 9 - 9x \hspace{1.0cm}\Rightarrow x = \frac{9}{13}$

$\displaystyle \text{When } y = \frac{2}{3} \hspace{1.0cm}\Rightarrow \sqrt{\frac{x}{1-x}} = \frac{2}{3} \hspace{1.0cm}\Rightarrow \frac{x}{1-x} = \frac{4}{9}$

$\displaystyle \Rightarrow 9x = 4 - 4x \hspace{1.0cm}\Rightarrow x = \frac{4}{13}$

$\displaystyle \text{Therefore the required solution is } : \frac{9}{13}, \frac{4}{13}$

$\\$

Question 18: Find the solution set of the equation $3x^2 - 8x - 3 = 0;$ when

(i) $x \in Z$ (integers) (ii) $x \in Q$ ( Rational Numbers)

Answer:

$\displaystyle 3x^2 - 8x - 3 = 0$

$\displaystyle \Rightarrow 3x^2 - 9x + x - 3 = 0$

$\displaystyle \Rightarrow 3x( x-3) + 1( x-3) = 0$

$\displaystyle \Rightarrow (x-3)(3x+1) = 0$

$\displaystyle \Rightarrow x = 3 \text{ or } x = \frac{-1}{3}$

$\displaystyle \text{(i) when } x \in Z, \text{ the solution set } = \{ 3 \}$

$\displaystyle \text{(ii) when } x \in Q, \text{ the solution set } = \{ 3, -\frac{1}{3} \}$

$\\$

Question 19: Solve: $( 2x-3)^2 = 25$

Answer:

$( 2x-3)^2 = 25$

$\Rightarrow 4x^2 - 12 x + 9 - 25 = 0$

$\Rightarrow 4x^2 - 12 x - 16 = 0$

$\Rightarrow x^2 - 3x - 4 = 0$

$\Rightarrow (x-4)(x+1) = 0$

$\Rightarrow x = 4 \text{ or } x = -1$

$\\$

$\displaystyle \text{Question 20: Solve for x } : 4 \Big( x - \frac{1}{x} \Big)^2 + 8 \Big(x + \frac{1}{x} \Big) = 29 , x \neq 0$

Answer:

$\displaystyle \text{Let } x + \frac{1}{x} = y$

$\displaystyle \Big( x - \frac{1}{x} \Big)^2 = \Big( x + \frac{1}{x} \Big)^2 - 4 = y^2 - 4$

$\displaystyle 4 \Big( x - \frac{1}{x} \Big)^2 + 8 \Big(x + \frac{1}{x} \Big) = 29$

$\displaystyle \Rightarrow 4(y^2 -4) + 8y = 29$

$\displaystyle \Rightarrow 4y^2 - 16 + 8y = 29$

$\displaystyle \Rightarrow 4y^2 + 8y - 45 = 0$

$\displaystyle \Rightarrow ( 2y+9)(2y-5) = 0$

$\displaystyle \Rightarrow y = \frac{-9}{2} \text{ or } y = \frac{5}{2}$

$\displaystyle \text{When } y = \frac{-9}{2} \hspace{1.0cm} \Rightarrow x + \frac{1}{x} = \frac{-9}{2} \Rightarrow2x^2 + 9x + 2 = 0$

$\displaystyle x = \frac{-9 \pm \sqrt{(9)^2 - 4(2)(2)}}{2(2)} = \frac{-9 \pm \sqrt{65}}{4}$

$\displaystyle \text{When } y = \frac{5}{2} \hspace{0.5cm} \Rightarrow x + \frac{1}{x} = \frac{5}{2} \hspace{0.5cm} \Rightarrow2x^2 - 5x + 2 = 0 \hspace{0.5cm} \Rightarrow (x-2)(2x-1) = 0 \\ \\ \Rightarrow x = 2 \text{ or } x = \frac{1}{2}$