Question

One of the roots of the quadratic equation x2−5mx+6m2=0 is 36. Find the greatest possible value of the second root. Help needed as soon as possible, Thank You.

Answer 1

The greatest possible value of the second root, β = 54

Step-by-step explanation:

The given quadratic equation:

$x^2-5mx+6m^2=0$

Let α and β be the roots of the given quadratic equation.

α = 36

To find,  the greatest possible value of the second root ( β) = ?

∴ The sum of the roots,

α + β = $\dfrac{-b}{a}$

⇒ 36 + β = $\dfrac{-(-5m)}{1}$

⇒ 5m = 36 + β               ............. (1)

The product of the roots,

α.β = $\dfrac{c}{a}$

⇒ $36.\beta=\dfrac{6m^2}{1}$

⇒ $6.\beta=m^2$                   ............. (2)        

From equations (1) and (2), we get  

⇒ $(\dfrac{36+\beta}{5})^{2}=6\beta$

⇒ $\beta^2-78\beta+1296=0$

⇒ $\beta^2-54B\beta-24B\beta+1296=0$

⇒ β(β - 54) - 24(β - 54) = 0

⇒ (β - 54)(β - 24) = 0

⇒ β - 54 = 0 or, β - 24 = 0

⇒ β = 54 or, β = 24

∴  The greatest possible value of the second root, β = 54