The greatest possible value of the second root, β = 54
Step-by-step explanation:
The given quadratic equation:
[tex]x^2-5mx+6m^2=0[/tex]
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 = [tex]\dfrac{-b}{a}[/tex]
⇒ 36 + β = [tex]\dfrac{-(-5m)}{1}[/tex]
⇒ 5m = 36 + β ............. (1)
The product of the roots,
α.β = [tex]\dfrac{c}{a}[/tex]
⇒ [tex]36.\beta=\dfrac{6m^2}{1}[/tex]
⇒ [tex]6.\beta=m^2[/tex] ............. (2)
From equations (1) and (2), we get
⇒ [tex](\dfrac{36+\beta}{5})^{2}=6\beta[/tex]
⇒ [tex]\beta^2-78\beta+1296=0[/tex]
⇒ [tex]\beta^2-54B\beta-24B\beta+1296=0[/tex]
⇒ 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
