Question

A quadratic equation x²+2(3x+p)=0 where p is a constant, has roots a and 2a, where a≠0

(a) find the value of a and of p
(b) hence, from a quadratic equation whose roots are a+1 and a-5


HELP ME PLISS​

Answer 1

see explanation

Explanation:

Since x = a and x = 2a are roots, then substitute these values into the equation.

x = a

a² + 2(3a + p) = 0, that is

a² + 6a + 2p = 0 → (1)

x = 2a

(2a)² + 2(3(2a) + p) = 0

4a² + 12a + 2p = 0 → (2)

Solving simultaneously

Subtract (1) from (2) term by term to eliminate p

3a² + 6a = 0 ← in standard form

3a(a + 2) = 0 ← in factored form

Equate each factor to zero and solve for a

3a = 0 ⇒ a = 0

a + 2 = 0 ⇒ a = - 2

Since a ≠ 0 then a = - 2

Substitute a = - 2 into either of the 2 equations and solve for p

substituting into (1)

(- 2)² + 6(- 2) + 2p = 0

4 - 12 + 2p = 0

- 8 + 2p = 0 ( add 8 to both sides )

2p = 8 ( divide both sides by 2 )

p = 4

Thus a = - 2 and p = 4

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a + 1 = - 2 + 1 = - 1

a - 5 = - 2 - 5 = - 7

If x = - 1 and x = - 7 are roots then the corresponding factors are

(x + 1) and (x + 7) and the quadratic equation is the product of the factors, that is

(x + 1)(x + 7) = 0 ← expand left side using FOIL

x² + 8x + 7 = 0

Answer 2

a = -2 and p = 4.

Explanation:

x²+2(3x+p)=0

x^2 + 6x + 2p = 0

For the general equation ax^2 + bx + c = 0, the sum of roots = -b/a and their product = c / a.

So here we have:

Sum of the roots:

a + 2a = -6/1 = -6

Product of the roots = 2a^2 = 2p/1 = 2p..

So 3a = -6 and

a = -2.

and

2(-2)^2 = 2p

8 = 2p

p = (-2)^2 = 4.