FIND QUADRATIC EQUATION WHEN ROOTS ARE GIVEN

Form a quadratic equation whose roots are

(i)  3, 4

(ii) 3+√7, 3-√7

(iii) (4+√7)/2 , (4-√7)/2

Question 1 :

3, 4

Solution :

α = 3, β = 4

x²-(α+β)x+αβ  =  0

Sum of roots :

α+β  =  3+4 ==> 7

Product of roots :

αβ  =  3(4) ==> 12

By applying those values in the general form we get,

x²-7x+12  =  0

Question 2 :

3 + √7 , 3 - √7

Solution :

α  =  3+√7, β = 3 - √7

Sum of roots :

α+β  =  3+√7+3-√7 

  =  6

Product of roots :

α β  =  (3+√7)(3-√7)

=  3² - 7

=  9 - 7

= 2

By applying those values in the general form we get,

x²-6x+2  =  0

Question 3 :

(4+√7)/2 , (4-√7)/2

Solution :

α = (4 + √7)/2, β = (4 - √7)/2

x²-(α+β)x+αβ  =  0

Sum of roots :

α + β  =  (4 + √7)/2 + (4 - √7)/2

=  (4 + √7 + 4 - √7)/2

=  8/2

  =  4

Product of roots :

α β = [(4 + √7)/2] [(4 - √7)/2] 

= (4² - (√7)²)/4

= (16 - 7)/4

= 9/4

by applying those values in the general form we get,

x²-4x+(9/4)  =  0

4x²-16x+9  =  0

Question 4 :

If the sum of product of a quadratic equation are -4 and -12 then quadratic equation is.

Solution :

Sum of roots (α + β) = -4

Product of roots (α β) = -12

x²-(-4)x + (-12) = 0

x²+ 4x - 12 = 0

Question 5 :

The roots of the quadratic equation 3x² - 2√6x + 2 = 0 are ____?

Solution :

3x² - 2√6x + 2 = 0

3x² - √6x - √6x + 2 = 0

√3 √3 x² - √3 √2x - √3 √2x + √2√2 = 0

√3x(√3x - √2) - √2(√3x - √2) = 0

(√3x - √2) (√3x - √2) = 0

Equating each factor to 0, we get

So, the roots are √2/√3 and √2/√3.

Question 6 :

If -5 is the root of the quadratic equation 2x² + px - 15 = 0 and the equation P(x² + x) + k = 0 has equal roots, then find values of p and k.

Solution :

2x² + px - 15 = 0

Since -5 is one of the roots of the quadratic equation, applying the value of x as -5, we get

2(-5)² + p(-5) - 15 = 0

2(25) - 5p - 15 = 0

50 - 15 - 5p = 0

-5p + 35 = 0

-5p = -35

Dividing by -5 on both sides, we get

p = 7

By applying the value of p in the equation, we get

2x² + 7x - 15 = 0

Solving by factoring,

2x² + 10x - 3x - 15 = 0

2x(x + 5) - 3(x + 5) = 0

(2x - 3) (x + 5) = 0

Equating each factor to 0, we get

x = 3/2 and x = -5

The roots of the equation is 

x = 3/2 and x = -5

P(x² + x) + k = 0

Since this equation has equal roots, then 

b² - 4ac = 0

a = P, b = P and c = k

P² - 4(P)(k) = 0

P² - 4Pk = 0

Applying the value of P, we get

7²- 4(7)k = 0

49 - 28k = 0

28k = 49

k = 49/28

k = 7/4

So, the value of P is 7 and value of k is 7/4.

Question 7 :

The value of k is ___, if 2 is the root of the following quadratic equation x² - (k + 1)x + k = 0.

Solution :

x² - (k + 1)x + k = 0

Since -2 is the root of the equation, applying the value of x as -2, we get

(-2)² - (k + 1)(-2) + k = 0

4 + 2k - 2 + k = 0

2 + 3k = 0

3k = -2

k = -2/3

So, the value of k is -2/3.

Question 8 :

If 2x² - (a + 6)2x + 12a = 0, then the roots are

Solution :

2x² - (a + 6)2x + 12a = 0

2x² - 2ax - 12x + 12a = 0

Factoring 2x from first two terms, we get

2x(x - a) - 12x + 12a = 0

Factoring 12 from third and fourth term, we get

2x(x - a) - 12(x - a) = 0

(2x - 12) (x - a) = 0

Equation each factor to 0, we get

2x = 12 and x = a

x = 6 and x = a

So, the roots are a and 6.

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