EXAMPLES ON QUADRATIC FUNCTIONS

About "Examples on quadratic functions"

Examples on quadratic functions :

Here we are going to see some examples on quadratic functions.

Question 1 :

Construct a quadratic equation with roots 7 and −3.

Solution :

Let α and β be the roots of the required quadratic equation

α  =  7,  β  =  -3

General form of a quadratic quadratic equation :

x2 - (α + β)x + α β  =  0

x2 - (7 + (-3))x + 7(-3)  =  0

x2 - 4x - 21  =  0

Question 2 :

A quadratic polynomial has one of its zeros 1 + √5 and it satisfies p(1) = 2. Find the quadratic polynomial.

Solution :

Let α  =  1 + √5, β be the roots of the required quadratic equation

p(x)  =  x2 - (1 + √5 + β)x + (1 + √5)β

p(1)  =  12 - (1 + √5 + β)1 + (1 + √5)β

2  =  1 - 1 - √5 - β + β + √5 β

2 + √5  =  √5 β

β  =  (2 + √5)/√5

 α + β  =  (1 + √5) + (2 + √5)/√5

  =  [√5(1 + √5) + (2 + √5)]/√5

  =  (√5 + 5 + 2 + √5)/√5

  =  (7 + 2√5)/√5

  =  (7√5 + 10)/5

  αβ  =  (1 + √5) ⋅ (2 + √5)/√5

Question 3 :

If α and β are the roots of the quadratic equation x2 + √2x + 3 = 0, form a quadratic polynomial with zeroes 1/α, 1/β .

Solution :

General form of a quadratic quadratic when α and β are the roots of the equation  :

x2 - (α + β)x + α β  =  0

Now, let us find sum and product of roots of the quadratic equation

x2 + √2x + 3 = 0

α + β  =  -√2/1  =  - √2

α β   =  3/1  =  3

here α  =  1/α and β  =  1/β

x2 - (1/α + 1/β)x + (1/α)(1/β)  =  0

x2 - ((α + β)/α β)x + (1/αβ)  =  0

x2 - ((√2)/3)x + (1/3)  =  0

3x2 + √2x + 1  =  0

Hence the required quadratic equation is 3x2 + √2x + 1  =  0.

Question 4 :

If one root of k(x − 1)2 = 5x − 7 is double the other root, show that k = 2 or −25.

Solution :

k(x − 1)2  =  5x − 7

k(x2 - 2x + 1)  =  5x − 7

kx2- 2kx - 5x + k + 7  =  0

kx2- x(2k + 5) + (k + 7)  =  0

If one root is α, then the other root β  =  2α

α + β  =  -b/a

α + 2α  =  (2k + 5)/k

3α  =  (2k + 5)/k

α  =  (2k + 5)/3k  ----(1)

αβ  =  c/a

α (2α)  =  (k + 7)/k

2α2  =  (k + 7)/k  --(2)

Applying (1) in the second equation, we get

2((2k + 5)/3k)2  =  (k + 7)/k 

(8K2 + 40K + 50) / 9k  =  (k + 7)/k 

(8K2 + 40K + 50)/9K  =  (k + 7)

8K2 + 40K + 50  =  9K(K + 7)

8K2 + 40K + 50  =  9K2 + 63k

9K2 - 8K2 + 63k - 40k - 50  =  0

K2+ 23k - 50  =  0

(k + 25) (k - 2)  =  0

k + 25  = 0       k - 2  =  0

k  =  -25            k  =  2

After having gone through the stuff given above, we hope that the students would have understood "Examples on quadratic functions"

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