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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 =

𝒙𝟐+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

𝒙𝟐 = 2 =

𝒙𝟒+𝟏

𝒙𝟐 = 𝒙𝟒 + 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) 𝒙𝟐 + 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

2x = 7, where, k = -1 (d)𝑥3-x2 = (x-1)3

Solution:

Clearly (d) holds.

Sol. Clearly (d) holds.