EQUATIONS OF THE FORM -

Equations of The Form- √(ax + b) = (cx + d)

We find those solutions only for which (ax + b) ≥ 0 and (cx + b) ≥ 0

Those values of ‘x’ which do not satisfy these conditions are called extraneous values. Such values are rejected. This is to be noted that, only positive value of the square root.

Example.1) Solve  √(2x + 7) = x + 2

Ans.)  √(2x + 7) = x + 2  ……………..(i)

=>   2x + 7 = (x + 2)²          [On squaring both sides of (i)]

=>   2x + 7 = x² + 4x + 4

=>   x²+ 2x – 3 = 0

In above given equation is in form of ax²+ bx + c = 0, here a = 1, b = 2, and c = (-)3. Now we have to factorize c = 3 in two part. So, the process is to multiply a X c = 1 X 3 = 3, then we have to do simple factorization of 3, and we find 3 = 3 X 1. So, 3 is the product of common factors 3, 1,  and hence we have to find out addend of 3. After bx there is negative (-) or minus (-) sign, so two obtained factor must be in subtracted form. Addend of 3, one must be 3, and another must be 1. So, two addend of 3 must be 3 & 1.

=>   x² + (3 – 1) x – 3 = 0

=>   x² + 3x – x – 3 = 0

=>   x (x + 3) – (x + 3) = 0

=>   (x + 3) (x – 1) = 0                   [By Factorization]

=>  (x + 3) = 0   or   (x – 1) = 0        [Zero Product Rule]

=>   x = - 3  or  x = 1

So, the solution set = {- 3, 1}

Checking – Putting x = -3 in (i) and we get –

L.H.S = √(2x + 7) = √[{2 X (-3)} + 7]

=  √(- 6 + 7)  = √1  =  1

R.H.S  =  x + 2  = - 3 + 2 = - 1

So, x = - 3 does not satisfy (i), so it is rejected.

Again we will consider, x = 1 in (i), and we get –

L.H.S = √(2x + 7) = √{(2 X 1) + 7}

= √(2 + 7) = √9 =  3

R.H.S =  x + 2  = 1 + 2 = 3

So, x = 1 satisfies (i)

Hence, solution set = {1}          (Ans.)

Equation Of the Form –  √(ax² + bx + c) = (dx + e)

We find those solutions for which ax² + bx + c ≥  0

The other values of ‘x’ will be extraneous. We reject them.

Caution – We shall take only positive value of the square root

Example)  Solve - √(2x² - 2x + 21) = (2x – 3)

Solution :  √(2x² - 2x + 21) = (2x – 3)  ………………….(i)

=>    (2x² - 2x + 21) = (2x – 3)²           [Squaring both side]

=>    2x² - 2x + 21 = 4x² - 12x + 9

=>    4x² - 12x + 9 – 2x² + 2x – 21 = 0

=>    2x² - 10x – 12 = 0

In above given equation is in form of ax²+ bx + c = 0, here a = 2, b = (-)10, and c = (-)12. Now we have to factorize c = 10 in two part. So, the process is to multiply a X c = 2 X 12 = 24, then we have to do simple factorization of 24, and we find 24 = 2 X 3 X 2 X 2. So, 24 is the product of common factors 2, 3, 2, 2, and hence we have to find out addend of 24. After bx there is negative (-) or minus (-) sign, so two obtained factor must be in subtracted form. Addend of 24, one must be 12, and another must be 2. So, two addend of 24 must be 12 & 2.

=>   2x² - (12 – 2)x – 12 = 0

=>   2x² - 12x + 2x – 12 = 0

=>   2x (x – 6) + 2 (x – 6) = 0

=>    (x – 6) (2x + 2) = 0                          [By Factorization]

=>    (x – 6) = 0   or  (2x + 2) = 0               [Zero Product Rule]

=>  x = 6  or  x = - 1

Checking, by putting x = 6 in (i), we get –

L.H.S  =  √(2x² - 2x + 21)

=  √{(2 X 6²) – (2 X 6) + 21}

=  √(72 – 12 + 21)

=  √81  =  √9² = 9

R.H.S = (2x – 3) = {(2 X 6) – 3}

= 12 – 3 =  9

Hence, x = 6 is a root of the given equation

Again, by putting x = - 1 in (i), and we get –

L.H.S  =  √(2x² - 2x + 21)

=  √[{2 X (- 1)²} – {2 X (- 1)} + 21]

=  √(2 + 2 + 21)

=  √25  =  √5² = 5

R.H.S = (2x – 3) = [{2 X (- 1)} – 3]

= - 2 – 3  =  - 5

So,     L.H.S ≠ R.H.S

Thus, x = - 1 does not satisfy (i)

So, it is an extraneous root

Hence, solution set = {6}      (Ans.)

Example.2) Find the quadratic equation whose solution set is {- 5, 7}

Ans.)  Since the solution set is {- 5, 7}, we have

x = - 5,  or  x = 7

=>  (x + 5) = 0  or  (x – 7) = 0

=>  (x + 5) (x – 7) = 0

=>  x² + 5x – 7x – 35 = 0        (Ans.)