If alpha, beta be the roots of the equation 2x2-3x+1=0 find an equatio

1.	If alpha, beta be the roots of the equation 2x2-3x+1=0 find an equation whose roots are alpha/2beta+3, beta/2alpha+3
Answer» Given equations is 2x² – 3x + 1 = 0 ⇒ 2x² – 2x – x + 1 = 0 ⇒ 2x(x – 1) – 1(x – 1) = 0 ⇒ (2x – 1)(x – 1) = 0 ⇒ 2x – 1 = 0 or x – 1 = 0 ⇒ x \(=\frac{1}{2}\) or x = 1 Hence, roots of given quadratic equation is 1 and \(\frac{1}{2}\). But given that \(\alpha\) and \(\beta\) are roots of given equation. Let \(\alpha \) = 1 and \(\beta=\frac{1}{2}\) Then \(\frac{\alpha}{2\beta}+3\) \(=\frac{1}{2\times \frac{1}{2}}+3\) = 1 + 3 = 4 And \(\frac{\beta}{2\alpha}+3\) \(=\frac{\frac{1}{2}}{2\times 1}+3\) \(=\frac{1}{4}+3\) \(=\frac{13}{4}\) The equations whose roots are 4 & \(\frac{13}{4}\) is (x – 4)(x – \(\frac{13}{4}\)) = 0 ⇒ x² – 4x – \(\frac{13}{4}\mathrm x\) + \(4\times \frac{13}{4}\) = 0 ⇒ x² – \(\frac{29}{4}\mathrm x\) + 13 = 0 Hence, the required equation is x² – \(\frac{29}{4}\mathrm x\) + 13 = 0.

If alpha, beta be the roots of the equation 2x2-3x+1=0 find an equation whose roots are alpha/2beta+3, beta/2alpha+3

Answer»

Given equations is 2x² – 3x + 1 = 0

⇒ 2x² – 2x – x + 1 = 0

⇒ 2x(x – 1) – 1(x – 1) = 0

⇒ (2x – 1)(x – 1) = 0

⇒ 2x – 1 = 0 or x – 1 = 0

⇒ x \(=\frac{1}{2}\) or x = 1

Hence, roots of given quadratic equation is 1 and \(\frac{1}{2}\).

But given that \(\alpha\) and \(\beta\) are roots of given equation.

Let \(\alpha \) = 1 and \(\beta=\frac{1}{2}\)

Then \(\frac{\alpha}{2\beta}+3\) \(=\frac{1}{2\times \frac{1}{2}}+3\) = 1 + 3 = 4

And \(\frac{\beta}{2\alpha}+3\) \(=\frac{\frac{1}{2}}{2\times 1}+3\) \(=\frac{1}{4}+3\) \(=\frac{13}{4}\)

The equations whose roots are 4 & \(\frac{13}{4}\) is

(x – 4)(x – \(\frac{13}{4}\)) = 0

⇒ x² – 4x – \(\frac{13}{4}\mathrm x\) + \(4\times \frac{13}{4}\) = 0

⇒ x² – \(\frac{29}{4}\mathrm x\) + 13 = 0

Hence, the required equation is x² – \(\frac{29}{4}\mathrm x\) + 13 = 0.