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Math Homework Help | Math Homework Help Service
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Sample of Math Homework Help Illustrations and Solutions:
Illustration: 1 Check whether the following are quadratic equations:
(i)x(x +1) + 8 = (x +2) (x+3) (ii) x (x+6) = 0 (iii) x +
1
𝑥
=1, x ≠ 0
(iv) 4x2 + 6x + 1 =x2
+ 2x + 3 (v) x2 +
1
𝑥2 = 2, x ≠ 0
Solution:
(i) x (x+1) + 8 = (x+2) (x+3)
= x2
+ (x+1) + 8 = x2
+2x +3x + 6
= 4x -2 = 0 = 2x -1 = 0, which is linear, ∴ given equation is not quadratic.
(ii) x(x+6) =x2
+ 6x = 0, which is a quadric equation
(iii) X +
𝟏
𝒙
= 1 =
𝒙 𝟐+𝟏
𝒙
= 1 = 𝑥2
-x +1= 0, which is a quadratic equation
(iv) 4x2 + 6x + 1 = 𝑥2
+ 2x + 3 = 4x2
–x2
+ 6x -2x + 1-3 =0, =3x2 + 4x -2 =0 which is
a quadratic equation
(v) x2
+
𝟏
𝒙 𝟐 = 2 =
𝒙 𝟒+𝟏
𝒙 𝟐 = 𝒙 𝟒
+ 1 = 2x2
= x4
-2x2
+ 1 =0.

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Which is not a quadratic equation (∴ 𝑥4
+2𝑥2
+ 1 is a polynomial of degree 4)
Illustration: 2 For the quadratic equation, determine whether the given value of x is
a solution of the given quadratic equation or not. (i) 3x2-x-1 =0 ; x = -1 (ii) 6x2 +
7x-5 = 0; x =
𝟏
𝟐
(iii) x2+2x-2 = 0 ; x = 3 – 1 and 3 + 1
Solution:
Putting x = -1 on the left hand side of the given equation, we get
L.H.S = 3(−1)2
- (-1) = 3 + 1-1 =3 ≠ 0 = 3𝑋2
+4X -2 =0
∴ x = -1 is not a root of the given equation
(ii) Putting x =
1
2
on the left hand side of the left hand side of the given equation, we get
L.H.S = 6
𝟏
𝟐
2 + 7
1
2
-5 = 6
𝟏
𝟒
+
𝟕
𝟐
- 5 =
𝟔+𝟏𝟒−𝟐𝟎
𝟒
= 0 = R.H.S.
∴ x =
1
2
is a root of the given equation.
(iii) Putting x = 3 -1 on the left hand side of the given equation, we get
L.H.S = ( 3 – 1)2
+ 2 3 − 1 -2 = 3 +1 -2 3 + 2 3 -2 -2
= 4 -4 = 0 = R.H.S ∴ x = 3 + 1 is not a root of the given equation.

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Again putting x = 3 + 1 on the left hand side of the given equation, we get
L.H.S =( 3 + 1)2
+ 2 3 + 1 -2 = 3 +1 +2 3 + 2 3 -2 -2
= 4 + 4 3 ≠ 0 ∴ x = 3 + 1 is not a root of the given equation.
Illustration: 3 Find the value of k so that x = 2 is root of the quadratic equation
3x2
-kx-2 = 0
(A.I.C.B.S.E.2002)
Solution:
Since 2 is root of the equation 3x2
–kx= 0 ∴ 3(2)2
- k(2)-2 =0
= 12-2k-2 = 0 = 2k = 10 and k =5
Illustration: 4 If one root of the equation 2x2
+ px – 15 = 0 is -5, find the value of p.
( C.B.S.E 2012)
Solution:
Since 2 is root of the equation 2x2
+ px – 15 = 0 ∴ 2 (-5)2
+ p (-5) -15 = 0= 50 – 5p –
15 = 0 = 5p = 7

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Illustration:5 Check, whether 3 is root of the equation.
𝑥2 − 5𝑥 + 6 + 𝑥2 − 9 = 4𝑥2 − 14𝑥 + 16
Solution:
Putting x = 3, on the L.H.S, we get
32 −5(3) + 6 + 32 − 9 = 9 − 9 = 0 + 0 = 0
R.H.S = 3(3)2 − 14(3) + 16 = 36 − 42 + 16 = 62 − 42 = 10
∴ L.H.S. ≠ R.H.S. ∴ x = 3 is not a root of the given equation.
Illustration: 6 Which of the following is not a quadratic equation ?
(a) X2
+
1
𝑥
= 1, x ≠ 0 (b) x +
1
𝑥
= 1,x ≠ 0 (c) x2
– 6x – 4 = 0 (d) x2
– 8 = 0
Solution:
(a) 𝒙 𝟐
+
𝟏
𝒙
= 1, = x3
+ 1 =x = x3
– x + 1 = 0
This is not a quadratic equation. [∵ x3
–x + 1 is a polynomial of degree 3]
(b) x +
1
𝑥
= 1 = x2
–x + 1 = 0 which is a quadratic equation.
(c) ,(d) are clearly quadratic equations. ∴ (a) holds.

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Illustration: 7 Which of the following is a quadratic equation.
(a) X2 + 2x + 1 = (4-x)2 + 3 (b) -2x2 = (5-x)
(c)(K+1) x2
+
3
2
x = 7, where, k = -1 (d)𝑥3
-x2
= (x-1)3
Solution:
Clearly (d) holds.
Sol. Clearly (d) holds.