Class 10: Quadratic Equations – Exercise 7(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Solve each of the following equations for $\displaystyle x$

Question 1: $\displaystyle x^2-8x+5=0$

Answer:

Comparing $\displaystyle x^2-8x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -8 \text{ and } c =5$

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

$\displaystyle \text{Therefore } x = \frac{-(-8) \pm \sqrt{(-8)^2-4(1)(5)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 7.32, 0.68$

$\displaystyle \\$

Question 2: $\displaystyle 5x^2+10x-3=0$

Answer:

Comparing $\displaystyle 5x^2+10x-3=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 5, b = 10 \text{ and } c =-3$

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

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

$\displaystyle \text{Solving we get } x = 0.26, -2.26$

$\displaystyle \\$

Question 3: $\displaystyle 2x^2-10x+5=0$

Answer:

Comparing $\displaystyle 2x^2-10x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = -10 \text{ and } c =5$

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

$\displaystyle \text{Therefore } x = \frac{-(-10) \pm \sqrt{(-10)^2-4(2)(5)}}{2(2)}$

$\displaystyle \text{Solving we get } x = 4.44, 0.56$

$\displaystyle \\$

$\displaystyle \text{Question 4: } 4x+ \frac{6}{x} +13=0$

Answer:

$\displaystyle 4x+ \frac{6}{x} +13=0$ or

$\displaystyle 4x^2+13x+6=0$

Comparing $\displaystyle 2x^2-10x+5=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 4, b = 13 \text{ and } c =6$

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

$\displaystyle \text{Therefore } x = \frac{-(13) \pm \sqrt{(13)^2-4(4)(6)}}{2(4)}$

$\displaystyle \text{Solving we get } x = -0.56, -2.69$

$\displaystyle \\$

Question 5: $\displaystyle x^2-3x-9=0$ [2007]

Answer:

Comparing $\displaystyle x^2-3x-9=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =-9$

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

$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 4.85, -1.85$

$\displaystyle \\$

Question 6: $\displaystyle x^2-5x-10=0$ [2013]

Answer:

Comparing $\displaystyle x^2-5x-10=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -5 \text{ and } c =-10$

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

$\displaystyle \text{Therefore } x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 6.53, -1.53$

$\displaystyle \\$

Question 7: $\displaystyle 3x^2-12x-1=0$

Answer:

Comparing $\displaystyle 3x^2-12x-1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 3, b = -12 \text{ and } c =-1$

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

$\displaystyle \text{Therefore } x = \frac{-(-12) \pm \sqrt{(-12)^2-4(3)(-1)}}{2(3)}$

$\displaystyle \text{Solving we get } x = 4.082, -0.082$

$\displaystyle \\$

Question 8: $\displaystyle x^2-16x+6=0$

Answer:

Comparing $\displaystyle x^2-16x+6=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -16 \text{ and } c =6$

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

$\displaystyle \text{Therefore } x = \frac{-(-16) \pm \sqrt{(-16)^2-4(1)(6)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 15.616, 0.384$

$\displaystyle \\$

Question 9: $\displaystyle 2x^2+11x+4=0$

Answer:

Comparing $\displaystyle 2x^2+11x+4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = 11 \text{ and } c =4$

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

$\displaystyle \text{Therefore } x = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(2)}$

$\displaystyle \text{Solving we get } x = -0.392, -5.108$

$\displaystyle \\$

Question 10: $\displaystyle x^4-2x^2-3=0$

Answer:

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

$\displaystyle \text{Let } x^2 = y$

$\displaystyle \text{Therefore } y^2-2y-3=0$

Comparing $\displaystyle y^2-2y-3=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -2 \text{ and } c =-3$

$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$

$\displaystyle \text{Solving we get } y = 3, -1$

$\displaystyle \text{Therefore } x^2 = 3\text{ or } x^2 = -1$ (imaginary number).

$\displaystyle \text{Hence } x = 1.732, -1.732\text{ or } \pm\sqrt{3}$

$\displaystyle \\$

Question 11: $\displaystyle x^4-10x^2+9=0$

Answer:

$\displaystyle x^4-10x^2+9=0$

$\displaystyle \text{Let } y = x^2$

Therefore we have $\displaystyle y^2-10y+9=0$

Comparing $\displaystyle y^2-10y+9=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -10 \text{ and } c =9$

$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } 7 = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$

$\displaystyle \text{Solving we get } y = 9, 1$

$\displaystyle \text{Therefore } x^2 = 9 \Rightarrow x = \pm 3\text{ or } x^2 = 1 \Rightarrow x = \pm 1$ .

$\displaystyle \text{Hence } x = \pm 3, \pm 1$

$\displaystyle \\$

Question 12: $\displaystyle (x^2-x)^2+5(x^2-x)+4=0$

Answer:

$\displaystyle \text{Let } (x^2-x)=y$

$\displaystyle \text{Therefore } y^2+5y+4=0$

Comparing $\displaystyle y^2+5y+4=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = 5 \text{ and } c =4$

$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } y = \frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(1)}$

$\displaystyle \text{Solving we get } y = -1, -4$

$\displaystyle \text{When } y = -1 \Rightarrow x^2-x+1=0$

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

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

$\displaystyle \text{Hence } x = Imaginary$

$\displaystyle \text{When } y = -4 \Rightarrow x^2-x+4=0$

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

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

$\displaystyle \text{Hence } x = Imaginary$

$\displaystyle \\$

Question 13: $\displaystyle (x^2-3x)^2-16(x^2-3x)-36=0$

Answer:

$\displaystyle \text{Let } (x^2-3x)=y$

$\displaystyle \text{Therefore } y^2-16y-36=0$

Comparing $\displaystyle y^2-16y-36=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 1, b = -16 \text{ and } c =-36$

$\displaystyle \text{Since } y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$\displaystyle \text{Therefore } y = \frac{-(16) \pm \sqrt{(-16)^2-4(1)(-36)}}{2(1)}$

$\displaystyle \text{Solving we get } y = 18, -2$

$\displaystyle \text{When } y = 18 \Rightarrow x^2-3x-18=0$

Comparing $\displaystyle x^2-3x-18=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =-18$

$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-18)}}{2(1)}$

$\displaystyle \text{Hence } x = 6, -3$

$\displaystyle \text{When } y = -2 \Rightarrow x^2-x+2=0$

Comparing $\displaystyle x^2-3x+2=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -3 \text{ and } c =2$

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

$\displaystyle \text{Hence } x = 2, 1$

$\displaystyle \text{Hence } x = 1, 2, 6, -3$

$\displaystyle \\$

$\displaystyle \text{Question 14: } \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}$

Answer:

$\displaystyle \sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}} = \frac{5}{2}$

$\displaystyle \text{Let } \sqrt{\frac{x}{x-3}} = y$

Therefore the equation becomes

$\displaystyle y + \frac{1}{y} = \frac{5}{2}$

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

Comparing $\displaystyle 2y^2-5y+2=0$ with $\displaystyle ay^2+by+c=0$ , we get $\displaystyle a = 2, b = -5 \text{ and } c =2$

$\displaystyle \text{Therefore } y = \frac{-(-3) \pm \sqrt{(-3)^2-4(1)(c)}}{2(2)}$

$\displaystyle \text{Hence } y = 2, 0.5$

When y = 2

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

Squaring both sides

$\displaystyle \frac{x}{x-3} = 4 \Rightarrow 3x=12 \Rightarrow x = 4$

Similarly, when y = 0.5

$\displaystyle \sqrt{\frac{x}{x-3}} = 0.25 \Rightarrow 3x=-3 \Rightarrow x = -1$

$\displaystyle \text{Hence } x = -1, 4$

$\displaystyle \\$

$\displaystyle \text{Question 15: } ( \frac{2x-3}{x-1} )-4( \frac{x-1}{2x-3} )=3$

Answer:

$\displaystyle ( \frac{2x-3}{x-1} )-4( \frac{x-1}{2x-3} )=3$

$\displaystyle \text{Let } ( \frac{2x-3}{x-1} ) = y$

$\displaystyle \text{Therefore } y - \frac{4}{y} =3$

$\displaystyle y^2-3y-4=0$

$\displaystyle y^2-4y+y-4=0$

$\displaystyle y(y-4)+(y-4)=0$

$\displaystyle (y-4)(y+1)=0 \Rightarrow y = 4, -1$

$\displaystyle \text{When } y = 4$

$\displaystyle \frac{2x-3}{x-1} =4$

$\displaystyle 2x-3=4x-4 \Rightarrow x = 0.5$

$\displaystyle \text{When } y = -1$

$\displaystyle \frac{2x-3}{x-1} =-1$

$\displaystyle 2x-3=-x+1 \Rightarrow x = \frac{4}{3}$

$\displaystyle \text{Hence } x = 0.5, \frac{4}{3}$

$\displaystyle \\$

$\displaystyle \text{Question 16: } ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}$

Answer:

$\displaystyle ( \frac{3x+1}{x+1} )+( \frac{x+1}{3x+1} )= \frac{5}{2}$

$\displaystyle \text{Let } ( \frac{3x+1}{x+1} ) = y$

$\displaystyle y + \frac{1}{y} = \frac{5}{2}$

$\displaystyle 2(y^2+1)=5y$

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

$\displaystyle 2y^2-4y-y+2=0$

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

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

$\displaystyle y = 2, 0.5$

$\displaystyle \text{When } y = 2$

$\displaystyle ( \frac{3x+1}{x+1} ) = 2$

$\displaystyle 3x+1=2x+2 \Rightarrow x =1$

$\displaystyle \text{When } y =0.5$

$\displaystyle ( \frac{3x+1}{x+1} ) = y$

$\displaystyle 6x+2=x+1 \Rightarrow x = - \frac{1}{5}$

$\displaystyle \text{Hence } x = 1, - \frac{1}{5}$

$\displaystyle \\$

$\displaystyle \text{Question 17: } 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10$

Answer:

$\displaystyle 3\sqrt{\frac{x}{5}} +3\sqrt{\frac{5}{x}} =10$

$\displaystyle \text{Let } \sqrt{\frac{x}{5}} = y$

$\displaystyle 3y+ \frac{3}{y} =10$

$\displaystyle 3y^2-10y+3=0$

$\displaystyle 3y^2-9y-y+3=0$

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

$\displaystyle (y-3)(3y-1)=0 \Rightarrow y = 3, \frac{1}{3}$

$\displaystyle \text{When } y = 3$

$\displaystyle \sqrt{\frac{x}{5}} = 3$

$\displaystyle \frac{x}{5} =9 \Rightarrow x = 45$

$\displaystyle \text{When } y = \frac{1}{3}$

$\displaystyle \sqrt{\frac{x}{5}} = \frac{1}{3}$

$\displaystyle \frac{x}{5} = \frac{1}{9} \Rightarrow x = \frac{5}{9}$

$\displaystyle \text{Hence } x = 45, \frac{5}{9}$

$\displaystyle \\$

$\displaystyle \text{Question 18: } 2x- \frac{1}{x} =7$ [2006]

Answer:

$\displaystyle 2x- \frac{1}{x} =7$

$\displaystyle 2x^2-7x-1=0$

Comparing $\displaystyle 2x^2-7x-1=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 2, b = -7 \text{ and } c =-1$

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

$\displaystyle \text{Therefore } x = \frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-1)}}{2(2)}$

$\displaystyle \text{Solving we get } x = 3.64, -0.14$

$\displaystyle \\$

Question 19: $\displaystyle 5x^2-3x-4=0$ [2012]

Answer:

Comparing $\displaystyle 5x^2-3x-4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 5, b = -3 \text{ and } c =-4$

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

$\displaystyle \text{Therefore } x = \frac{-(-3) \pm \sqrt{(-3)^2-4(5)(-4)}}{2(5)}$

$\displaystyle \text{Solving we get } x = 1.243, -0.643$

If we want only three significant figures than $\displaystyle x = 1.24, -0.643$

$\displaystyle \\$

Question 20: $\displaystyle (x-1)^2-3x+4=0$ [2014]

Answer:

$\displaystyle (x-1)^2-3x+4=0$

$\displaystyle x^2+1-2x-3x+4=0$

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

Comparing $\displaystyle 5x^2-3x-4=0$ with $\displaystyle ax^2+bx+c=0$ , we get $\displaystyle a = 1, b = -5 \text{ and } c =5$

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

$\displaystyle \text{Therefore } x = \frac{-(-5) \pm \sqrt{(-5)^2-4(1)(5)}}{2(1)}$

$\displaystyle \text{Solving we get } x = 3.618, 1.382$

If we want only two significant figures than $\displaystyle x = 3.6, 1.3$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 10: Quadratic Equations – Exercise 7(c)

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