+0

+2
213
1

$$Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0. Find \frac{1}{2a} + \frac{1}{2b}.$$

Feb 16, 2018

#1
+100169
+3

Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0. Find \frac{1}{2a} + \frac{1}{2b}.

$$\text{Let }\alpha\;\; and \;\ \beta\;\; \text{be the solutions of the quadratic equation }\\ 2x^2 - 8x + 7 = 0. \;\;\;Find \;\;\ \frac{1}{2a} + \frac{1}{2b}\\ \frac{1}{2\alpha} + \frac{1}{2\beta}=\frac{\beta + \alpha}{2\alpha\beta}$$

Now you can do this the long way and work out what the roots are but I expect you are supposed to know this:

$$\boxed{If \;\;ax^2+bx+c=0 \text{ and the roots are }\alpha \;\;and \;\; \beta\;\;then\\ \alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\}~\\$$

So for your question

$$\alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\ \alpha + \beta = \frac{8}{2}\;\;and \;\; \alpha \beta = \frac{7}{2}\\$$

$$\frac{1}{2\alpha} + \frac{1}{2\beta}\\=\frac{\beta + \alpha}{2\alpha\beta}\\ =\frac{8}{2}\div\frac{2\times7}{2}\\ =4\div7\\ =\frac{4}{7}$$

.
Feb 16, 2018

#1
+100169
+3

Let a and b be the solutions of the quadratic equation 2x^2 - 8x + 7 = 0. Find \frac{1}{2a} + \frac{1}{2b}.

$$\text{Let }\alpha\;\; and \;\ \beta\;\; \text{be the solutions of the quadratic equation }\\ 2x^2 - 8x + 7 = 0. \;\;\;Find \;\;\ \frac{1}{2a} + \frac{1}{2b}\\ \frac{1}{2\alpha} + \frac{1}{2\beta}=\frac{\beta + \alpha}{2\alpha\beta}$$

Now you can do this the long way and work out what the roots are but I expect you are supposed to know this:

$$\boxed{If \;\;ax^2+bx+c=0 \text{ and the roots are }\alpha \;\;and \;\; \beta\;\;then\\ \alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\}~\\$$

So for your question

$$\alpha + \beta = \frac{-b}{a}\;\;and \;\; \alpha \beta = \frac{c}{a}\\ \alpha + \beta = \frac{8}{2}\;\;and \;\; \alpha \beta = \frac{7}{2}\\$$

$$\frac{1}{2\alpha} + \frac{1}{2\beta}\\=\frac{\beta + \alpha}{2\alpha\beta}\\ =\frac{8}{2}\div\frac{2\times7}{2}\\ =4\div7\\ =\frac{4}{7}$$

Melody Feb 16, 2018