If a and B are the zeros of the quadratic polynomial f(x) = x2- 5x + 4 find the value of 1/a+1/b-2ab

Step-by-step explanation:

Given the quadratic polynomial given as g(x) = x²- 5x + 4, the zeros of the quadratic polynomial occurs at g(x) = 0 such that x²- 5x + 4 = 0.

Factorizing the resulting equation to get the roots

x²- 5x + 4 = 0

(x²- x)-(4 x + 4) = 0

x(x-1)-4(x-1) = 0

(x-1)(x-4) = 0

x-1 = 0 and x-4 = 0

x = 1 and x = 4

Since a and b are known to be the root then we can say a = 1 and b =4

Substituting the given values into the equation  1/a+1/b-2 ab , we will have;

= 1/1 + 1/4 - 2*1*4

= 1 + 1/4 - 8

= 5/4 - 8

Find the Lowest common multiple

= (5-32)/4

= -27/4

Hence the required value is -27/4

answer:   the paragraph proof is mentioned below.

step-by-step explanation:

given: in a δxwy where

such that, ∠wzx ≅ ∠yzx and zw≅zy

we have to prove that: segment zx is a perpendicular bisector of segment wy.

prove: since, zw≅zy ⇒ ∠wzy is a supplementary angle.

then, ∠wzy= 180°

⇒∠wzx + ∠yzx = 180° ( because xz divides ∠wzy into two parts)

but, ∠wzx ≅ ∠yzx ( given)

⇒∠wzy+∠wzy=180° ( by substitution)

⇒2∠wzy=180°

⇒∠wzy=90°

⇒∠wzy=∠yzx =90°

therefore, xz⊥wy where wz=zy

⇒zx is a perpendicular bisector of wy.