FINDING FOR UNKNWONS FOR A QUADRATIC EQUATION WHEN ROOTS ARE EQUAL

Find the values of k for which the given quadratic equations have real and equal roots.

Problem 1 :

4x² + kx + 9 = 0

Solution :

Since the roots are real and equal, b² - 4ac = 0

From the equation given above,

a = 4, b = k and c = 9

k² - 4(4)(9) = 0

k² - 144 = 0

k² = 144

k = √144

k = ±12

Then the possible values of k are -12 and 12.

Problem 2 :

kx² - 5x + k = 0

Solution :

Since the roots are real and equal, b² - 4ac = 0

From the equation given above,

a = k, b = -5 and c = k

(-5)² - 4(k)(k) = 0

25 - 4k² = 0

4k² = 25

k² = 25/4

k = √(25/4)

k = ±5/2

Then, the values of k are -5/2 and 5/2.

Problem 3 :

x² - 7(3+k) + 4 - 2x(1 + k) = 0

Solution :

x² - 7(3+k) + 4 - 2x(1 + k) = 0

The given quadratic equation is not in the standard form, by arranging it.

x² - 2x(1 + k) - 7(3+k) + 4 = 0

From this equation, we get

a = 1, b = -2(1 + k) and c = - 7(3+k) + 4

c = -21 - 7k + 4

c = -17 - 7k

[-2(1 + k)]² - 4(1)(-17 - 7k) = 0

4(1 + k)² + 68 + 28k = 0

Using algebraic identity, finding the expansion

4(1 + 2k + k²) + 68 + 28k = 0

4 + 8k + 4k² + 68 + 28k = 0

4k² + 28k + 8k + 68 + 4 = 0

4k² + 36k + 72 = 0

Dividing the entire equation by 4, we get

k² + 9k + 18 = 0

Using the method of factoring, we can solve the equation above.

k² + 3k + 6k + 18 = 0

k(k + 3) + 6(k + 3) = 0

(k + 3) (k + 6) =0

Equating each factor to 0, we get

k = -3 and k = -6

Problem 4 :

Consider x² - 2x + m = 0. Find the discriminant, hence find the values of m for which the equation has

a) repeated roots

b) 2 distinct real roots

c)  no real roots

Solution :

x² - 2x + m = 0

a = 1, b = -2 and c = m

a) roots are repeated = roots are equal

Then b² - 4ac = 0

The given quadratic equation is not in the standard form, by arranging it.

(-2)² - 4(1)(m) = 0

4 - 4m = 0

4m = 4

m = 1

When m = 1, the quadratic equation will have repeated roots.

b) If the quadratic equation has two distinct real roots, then

b² - 4ac > 0

(-2)² - 4(1)(m) > 0

4 - 4m > 0

-4m > -4

Dividing by -4 on both sides, we get

m < 1

c) If the quadratic equation has no real roots, then

b² - 4ac < 0

(-2)² - 4(1)(m) < 0

4 - 4m < 0

-4m < -4

Dividing by -4 on both sides, we get

m > 1

Problem 5 :

If -4 is a root of the quadratic equation x² + px - 4 = 0 and the quadratic equation 12x² + 4px + k = 0 has equal roots, the find the value of k.

Solution :

For the quadratic equation, x² + px - 4 = 0 we have a root which is -4.

Let α and β be roots of the quadratic equation. Then α = -4

Sum of roots (α + β) = -b/a

Product of roots α β = c/a

a = 1, b = p and c = -4

α + β = -p/1 ==> -p 

α β = -4/1 ==> -4

-4 + β = -p-----(1)

-4β = -4

β = 1

Applying the α = -4 and β = 1 in (1), we get

-4 + 1 = -p

p = 3

Applying the value of p in the equation 12x² + 4px + k = 0

12x² + 4(3)x + k = 0

12x² + 12x + k = 0

Since the roots of this quadratic equation are equal, 

b² - 4ac = 0

a = 12, b = 12 and c = k

(12)² - 4(12)(k) = 0

144 - 48k = 0

-48k = -400

k = 400/48

k = 25/3

Problem 6 :

If the equation (1+k²)x² + 2kqx + (q² - p²) = 0 has equal roots, then shown that q² = p²(1 + k²).

Solution :

(1+k²)x² + 2kqx + (q² - p²) = 0

a = 1+k², b = 2kq and c = q² - p²

Since the roots of this quadratic equation are equal, 

b² - 4ac = 0

(2kq)² - 4(1+k²) (q² - p²) = 0

4k²q² - 4(q² - p² + k²q² - k²p²) = 0

4k²q² - 4q² + 4p² - 4k²q² + 4k²p² = 0

4p² + 4k²p² =  4q²

Dividing by 4, we get

p² + k²p² =  q²

Factoring p², we get

p² (1 + k²) =  q²

Hence proved