Question

I need a well explained answer


If x=1 & x=2 are the solutions of the equation ax^2+bx+1=0,then find the value of a &b​

Answer 1

Given :

• x = 1, x = 2 are the solutions of the equation ax²+bx+1=0

To find :

• VaIue of a and b

SoIution :

Given quadratic equation : ax² + bx + 1 = 0

Zeros of that quadratic equation are 1 and 2

Now we know ;

⇒ Sum of zeros = -b/a

⇒ 1 + 2 = -b/a

⇒ 3 = -b/a

⇒ - b = 3a

⇒ b = -3a ............. (i)

Now we know ;

⇒ Product of zeros = c/a

⇒ 1 × 2 = c/a

⇒ 2 = 1/a

⇒ 2a = 1

⇒ a = 1/2

Now putting vaIue of a in (i) :

⇒ b = -3 × 1/2

⇒ b = -3/2

Therefore,

VaIue of a and b are (1/2) and (-3/2) respectiveIy.

Answer 2

Given that,

x = 1 and x = 2.

Given equation, ax² + bx + 1 = 0

We've to find value of a and b.

Put x = 1 in given equation.

→ a(1)² + b(1) + 1 = 0

→ a + b = -1

→ a = - b - 1......(1)

put x = 2 in given equation.

→ a(2)² + b(2) + 1 = 0

→ a(4) + b(2) + 1 = 0

→ 2(2a + b) = -1

→ 2a + b = -1/2......(2)

Put value of (1) in (2).

→ 2(-b-1) + b = -1/2

→ -2b - 2 + b = -1/2

→ -b - 2 = -1/2

→ -b = -1/2 + 2

→ -b = (-2+4)/2

→ -b = -2/2

→ b = 1

Put the value of b in (1).

→ a + (1) = -1

→ a = -1 - 1

→ a = -2

Hence, value of a and b are -2 and 1 respectively.