Question

If x1 < x2 and x1, x2 are the roots of x2

– 26x + 120 = 0 then the value of
√x1+√x2​

Answer 1

√X+√X₂ = √26

Explanation:

Given Equation:

x2 – 26x + 120 = 0

To find:

The roots X₁ and X₂

Formula:

α  = $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

β = $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

From the equation,

a = coefficient of x²

b = coefficeint of x

c = constant

a = 1

b = -26

c = 120

α  = $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$

α  = $\frac{-(-26)+\sqrt{(-26)^ {2}-4(1)(120) } }{2(1)}$

α  = $\frac{26+\sqrt{676-4(120) } }{2}$

α  = $\frac{26+\sqrt{676-480 } }{2}$

α  = $\frac{26+\sqrt{196 } }{2}$

α  = $\frac{26+14 }{2}$

α  = $\frac{40}{2}$

α  = 20

β = $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

β = $\frac{-(-26)-\sqrt{(-26)^ {2}-4(1)(120) } }{2(1)}$

β  = $\frac{26-\sqrt{676-4(120) } }{2}$

β = $\frac{26-\sqrt{676-480 } }{2}$

β  = $\frac{26-\sqrt{196 } }{2}$

β = $\frac{26-14 }{2}$

β  = $\frac{12}{2}$

β = 6

The roots are 20 and 6

X₁ = 20

X₂ = 6

√X+√X₂ = √20 + √6

√X+√X₂ = √26