LEARN MATH STEP BY STEP THROUGH VERY EASY PROCESS

QUADRATIC EQUATION - QUADRATIC EQUATIONS OF THE FORM

__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.)**

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