Given below are the Quadratic Formula Worksheet with Answers Class 10 Maths
$ax^2 +bx +c =0$
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.$x = - 4, x = 4$

## Summary

This Quadratic Formula Worksheet Class 10 Maths is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.You can also download through below link

Go back to Class 10 Main Page using below links

### 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

Sound Class 8 Science Quiz

Limits Class 11 MCQ

Circles in Conic Sections Class 11 MCQ

Plant Kingdom free NEET mock tests

The Living World free NEET mock tests