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# help algebra

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The roots of 7x^2 + x - 5 = 0 are a and b. Compute 1/a + 1/b.

Jul 10, 2021

#1
+12378
+1

The roots of 7x^2 + x - 5 = 0 are a and b. Compute 1/a + 1/b.

Hello Guest!

$$7x^2 + x - 5 = 0\\ x^2+\frac{1}{7}x-\frac{5}{7}=0\\ x=-\frac{1}{14}\pm \sqrt{\frac{1}{196}+\frac{5}{7}}\\ x=-\frac{1}{14}\pm\sqrt{\frac{1+5\cdot 28}{196}}\\$$

$$x=-\frac{1}{14}\pm \frac{1}{14}\sqrt{141}\\ a=-\frac{1+\sqrt{3\cdot 47}}{14}\\ b=-\frac{1-\sqrt{3\cdot 47}}{14}\\$$

$$\frac{1}{a}+\frac{1}{b}=-\frac{14}{1+\sqrt{3\cdot 47} }-\frac{14}{1-\sqrt{3\cdot 47} }\\ =\frac{-14\cdot (1-\sqrt{3\cdot 47}+1+\sqrt{3\cdot 47})}{1-3\cdot 47}=\frac{-28}{-140}$$

$$\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{5}$$

!

Jul 11, 2021
#2
+26213
+2

The roots of $$7x^2 + x - 5 = 0$$ are $$a$$ and $$b$$.
Compute $$\dfrac{1}{a} + \dfrac{1}{b}$$
.

By Vieta:

$$\begin{array}{|rcll|} \hline 7x^2 + x - 5 &=& 0 \quad | \quad : 7 \\ x^2 + \underbrace{\frac{1}{7}}_{=-(a+b)}x \underbrace{-\frac{5}{7}}_{=ab} &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline a+b &=& -\dfrac{1}{7} \\\\ ab &=& -\dfrac{5}{7} \\\\ \hline \dfrac{a+b}{ab} &=& -\dfrac{1}{7} \over -\dfrac{5}{7} \\\\ \dfrac{a}{ab}+\dfrac{b}{ab} &=& \dfrac{1}{7} \over \dfrac{5}{7} \\\\ \dfrac{1}{a}+\dfrac{1}{b} &=& \dfrac{1}{7} \times \dfrac{7}{5} \\\\ \mathbf{\dfrac{1}{a}+\dfrac{1}{b}} &=& \mathbf{\dfrac{1}{5}} \\ \hline \end{array}$$

Jul 11, 2021
#3
+12378
0

Please, what does it say on Vieta, heureka?

asinus  Jul 11, 2021