Question

Find the values of a and b for which x = 3 / 4 and x = - 2 are the roots of the equation ax^2 + bx- 6 = 0.

Answer 1

Let the Given Quadratic Polynomial ax² + bx - 6 be P(x)

⇒ P(x) = ax² + bx - 6

Given that  $\frac{3}{4}$  is the Root of the Given Quadratic Equation

$\implies P(\frac{3}{4}) = 0$

$\implies a(\frac{3}{4})^2 + b(\frac{3}{4}) - 6 = 0$

$\implies a(\frac{9}{16}) + b(\frac{3}{4}) - 6 = 0$

$\implies 9a + 12b - 96 = 0$

$\implies9a + 12b = 96$

$\implies 3a + 4b = 32$ --------------- [1]

Given that -2 is the Root of the Given Quadratic Equation

⇒ P(-2) = 0

⇒ a(-2)² + b(-2) - 6 = 0

⇒ 4a - 2b = 6

Multiplying with 2 we get :

⇒ 8a - 4b = 12 ---------------- [2]

Adding Both Equation [1] and [2] we get :

⇒ 3a + 4b + 8a - 4b = 32 + 12

⇒ 11a = 44

⇒ a = 4

Substituting a = 4 in Equation [1] we get :

⇒ 3(4) +4b = 32

⇒ 12 + 4b = 32

⇒ 4b = 20

⇒ b = 5

So, The Values of a = 4 and b = 5

Given Equation is : 4x² + 5x - 6 = 0

Answer 2

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