$ax^2 +bx +c =0$
Where a, b and c are real numbers and a ≠0
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$

Question 1. State which all quadratic equations have real roots, no real roots
a. $x^2 + 8x+7=0$
b. $3x^2 +5x+1=0$
c. $9x^2 +x -3=0$
d. $11x^2 +5x+1=0$
e. $10x^2 +3x+7=0$
f. $2x^2 -6x+3=0$
g. $2x^2 -x-2=0$
Solution
Nature of roots of Quadratic equation
 S.no Condition Nature of roots 1 $b^2 -4ac > 0$ Two distinct real roots 2 $b^2-4ac =0$ One real root 3 $b^2-4ac < 0$ No real roots

Real roots: :(a), (b), (c),(d) ,(e),(f),(g)
No real roots : (d) ,(e)

Question 2. Find the roots of the quadratic equation using Quadratic Formula
a. $x^2-3x-10=0$
b. $x^2 -11x+30=0$
Solution
a.
$x^2-3x-10=0$
Comparing this to $ax^2 + bx +c=0$
We have a =1, b=-3 and c =-10
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
or $x=\frac {3 \pm \sqrt {9 +40}}{2}$
$x=\frac {3 \pm 7}{2}$
or
$x=\frac {3 + 7}{2}$ or $x=\frac {3 - 7}{2}$
So roots are x=5 and -2
b. $x^2 -11x+30=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=-11 and c =30
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
$x=\frac {11 \pm \sqrt {121 -120}}{2}$
$x=\frac {11 \pm 1}{2}$
or
$x=\frac {11 + 1}{2}$ or $x=\frac {11 - 1}{2}$
So
Roots are 5 and 6

Question 3. Find the roots of the quadratic equation using Quadratic formula
a. $x^2 +4x-5=0$
b. $2x^2-7x+3=0$

Solution
a.
$x^2 +4x-5=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=4 and c =-5
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=1 or -5
b.
$2x^2-7x+3=0$
Comparing this to $ax^2 + bx +c=0$
a=2, b=-7 and c =3
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=1/2 or 3

Question 4
Find the roots using Quadratic Formula
a. $x^2+ (7 - x)^2= 25$
b. $y^2+ (y+2)^2 =580$
c. $11x^2 -31x -6 =0$
d. $9 -y - 10y^2 =0$
e. $14x +4x^2 =2x -5$
f. $3y^2 + 4y=2(y+4)$
g. $2x^2 -5x +3 =0$
h. $\frac {x+1}{x-1} + \frac {x-2}{x+2} =3$ ,$x \neq 1$, $x \neq -2$
i. $x^2 + 5x =-1$
j. $\sqrt {2x + 9} + x=13$
k. $\frac {1}{a} + \frac {1}{b} + \frac {1}{x} = \frac {1}{a+b+x}$
l. $\sqrt {2} x^2 + 7x + 5 \sqrt {2}=0$
m. $\frac {3}{x+1} + \frac {4}{x-1} = \frac {29}{4x-1};x \neq 1,-1,1/4$
Solution
a.
$x^2+ (7 - x)^2= 25$
$x^2 + 49 +x^2 -14x =25$
$2x^2 -14x + 24=0$
$x^2 -7x + 12=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=-7 and c =12
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=3 or 4

b.
$y^2+ (y+2)^2 =580$
$y^2 + y^2 + 4 + 4y =580$
$2y^2 + 4y -576=0$
$y^2 + 2y-288=0$
Comparing this to $ay^2 + by +c=0$
a=1, b=2 and c =-288
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
y=16 or -18

c.
$11x^2 -31x -6 =0$
Comparing this to $ax^2 + bx +c=0$
a=11, b=-31 and c =-6
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
x=3 or -2/11

d.
$9 -y - 10y^2 =0$
$10y^2 + y -9=0$ Comparing this to $ay^2 + by +c=0$
a=10, b=1 and c =-9
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
So
y=-1 or 9/10

e.
$14x +4x^2 =2x -5$
$4x^2 + 12x +5=0$
Comparing this to $ax^2 + bx +c=0$
a=4, b=12 and c =5
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so x=-1/2 or -5/2

f.
$3y^2 + 4y=2(y+4)$
$3y^2 + 4y = 2y + 8$
$3y^2 + 2y -8=0$
Comparing this to $ay^2 + by +c=0$
a=3, b=2 and c =-8
$y= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so y=-2 or 4/3

g.
$2x^2 -5x +3 =0$
Comparing this to $ax^2 + bx +c=0$
a=2, b=-5 and c =3
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so y=3/2 or 1

h.
$\frac {x+1}{x-1} + \frac {x-2}{x+2} =3$
$\frac { (x+1)(x+2) + (x-2)(x-1)}{(x-1)(x+2)}=3$
$x^2 + 3x +2 + x^2 -3x + 2= 3(x^2 +x -2)$
$2x^2 +4 = 3x^2 +3x -6$
$x^2 +3x -10=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=3 and c =-10
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so x=-5 or 2

i.
$x^2 + 5x =-1$
$x^2 + 5x +1=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=5 and c =1
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
so,$x = \frac {-5 + \sqrt {21}}{2}$ or $x= \frac {-5 - \sqrt {21}}{2}$

j.
$\sqrt {2x + 9} + x=13$
$\sqrt {2x + 9} = 13-x$
Squaring both the sides $2x+9 = (13-x)^2$
$2x+ 9 = 169 + x^2 -26x$ $x^2 -28x +160=0$ Comparing this to $ax^2 + bx +c=0$
a=1, b=-28 and c =160
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
x=20,8
k.
$\frac {1}{a} + \frac {1}{b} + \frac {1}{x} = \frac {1}{a+b+x}$
$\frac {bx + ax + ab}{abx} =\frac {1}{a+b+x}$
$(bx + ax + ab)(a+b+x) = abx$
$abx + b^2x + bx^2 + a^2 x+ abx+ ax^2 + a^b + ab^2 + abx = abx$
$x^2(a + b) + x(a^2 + b^2 + 2ab) + ab(a+b) =0$
$x^2(a+b)+ x(a+b)^2 + ab(a+b) =0$
Dividing by (a+b)
$x^2 + x(a+b) + ab= 0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=(a+b) and c =ab
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
x=-a,-b

l.
$\sqrt {2} x^2 + 7x + 5 \sqrt {2}=0$
Comparing this to $ax^2 + bx +c=0$
$a=\sqrt {2}$, b=7 and $c =5 \sqrt {2}$
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
$x = \frac {-7 \pm \sqrt {7^2 -4 \times \sqrt {2} \times 5 \sqrt {2}}}{2\sqrt {2}}$
$x=\frac {-7 \pm \sqrt {49-40}}{2\sqrt {2}}$
or
$x= -\sqrt {2}$ or $x= -\frac {5\sqrt{2}}{2}$

m. $\frac {3}{x+1} + \frac {4}{x-1} = \frac {29}{4x-1};x \neq 1,-1,1/4$
$\frac {3(x-1) + 4(x+1)}{(x+1)(x-1)} = \frac {29}{4x-1}$
$\frac {7x +1}{x^2 -1} = \frac {29}{4x-1}$
$(7x+1)(4x-1) = 29(x^2-1)$
$28x^2 -7x+4x -1 = 29x^2 -29$
$x^2 + 3x-28=0$
Comparing this to $ax^2 + bx +c=0$
a=1, b=3 and c =-28
$x= \frac {-b \pm \sqrt {b^2 -4ac}}{2a}$
x=-7 or 4
Question 5
Solve the below quadratic equation using quadratic formula a. $x^{2} + 4 x = -1$
b. $3 x^{2} + 3 x - 1 = 0$
c. $3 x^{2} + 2 x = 2$
d. $2 x^{2} = 1 -3x$
e. $5 x^{2} = 2 (1-x)$
f. $4 x^{2} + 2 x = 0$
g. $4x( x+ 1) =1$
h. $\frac {1}{x+1} + \frac {2}{x+2} = \frac {4}{x+4},x \neq -1,-2,-4$
i $x^{2} + 2 x - 2 = 0$
j. $4 x^{2} + 4 x - 2 = 0$
k. $3 x^{2} + 4 x + 1 = 0$
l. $4 x^{2} + 4 x - 2=0$
m. $x^{2} + 3 x + 2 = 0$
n. $5 x^{2} + x = 0$
o. $\frac {16}{x} -1 = \frac {15}{x+1},x \neq 0,-1$
Solution
a. $x = -2 - \sqrt{3}, x = -2 + \sqrt{3}$
b. $x = - \frac{1}{2} + \frac{\sqrt{21}}{6}, x = - \frac{\sqrt{21}}{6} - \frac{1}{2}$
c. $x = - \frac{1}{3} + \frac{\sqrt{7}}{3}, x = - \frac{\sqrt{7}}{3} - \frac{1}{3}$
d. $x = - \frac{3}{4} + \frac{\sqrt{17}}{4}, x = - \frac{\sqrt{17}}{4} - \frac{3}{4}$
e. $x = - \frac{1}{5} + \frac{\sqrt{11}}{5}, x = - \frac{\sqrt{11}}{5} - \frac{1}{5}$
f. $x = - \frac{1}{2}, x = 0$
g. $x = - \frac{1}{2} + \frac{\sqrt{2}}{2}, x = - \frac{\sqrt{2}}{2} - \frac{1}{2}$
h. $x = 2 +2 \sqrt{3}, x = 2 - 2 \sqrt{3}$
i. $x = -1 + \sqrt{3}, x = - \sqrt{3} - 1$
j. $x = - \frac{1}{2} + \frac{\sqrt{3}}{2}, x = - \frac{\sqrt{3}}{2} - \frac{1}{2}$
k. $x = -1, x = - \frac{1}{3}$
l. $x = - \frac{1}{2} + \frac{\sqrt{3}}{2}, x = - \frac{\sqrt{3}}{2} - \frac{1}{2}$
m. $x = -2, x = -1$
n. $x = - \frac{1}{5}, x = 0$
o.$x = - 4, x = 4$

• Notes Assignments NCERT Solutions

Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

You can use above books for extra knowledge and practicing different questions.

### Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

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