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# can you help plz

0
365
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+187

Let $$p(x) = 2x^2 + 7x + c$$, where c is not equal to 0. If r1 and r2 are the roots of p(x), then find the value of 1/r1 + 1/r2 in terms of c.

Dec 9, 2018

#1
+6187
+1

$$\text{just use the quadratic formula}\\ r_1,r_2 = \dfrac{-7\pm\sqrt{49-8c}}{4}\\ \dfrac{1}{r_1}+\dfrac{1}{r_2} = \dfrac{4}{-7+\sqrt{49-8c}}+ \dfrac{4}{-7-\sqrt{49-8c}}=\\ \dfrac{-56}{49-49+8c} = \dfrac{-56}{-8c}= \dfrac{7}{c}$$

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Dec 9, 2018

#1
+6187
+1

$$\text{just use the quadratic formula}\\ r_1,r_2 = \dfrac{-7\pm\sqrt{49-8c}}{4}\\ \dfrac{1}{r_1}+\dfrac{1}{r_2} = \dfrac{4}{-7+\sqrt{49-8c}}+ \dfrac{4}{-7-\sqrt{49-8c}}=\\ \dfrac{-56}{49-49+8c} = \dfrac{-56}{-8c}= \dfrac{7}{c}$$

Rom Dec 9, 2018
#2
+24978
+10

Let $$\large{p(x) = 2x^2 + 7x + c}$$,
where $$c$$ is not equal to $$0$$.
If $$\large{r_1}$$ and $$\large{r_2}$$ are the roots of $$\large{p(x)}$$,

then find the value of $$\large{\dfrac{1}{r_1}} + \large{\dfrac{1}{r_2}}$$in terms of $$\large{c}$$.

$$\begin{array}{|rcll|} \hline p\left(\dfrac{1}{x}\right) = 2\left(\dfrac{1}{x}\right)^2 + 7\dfrac{1}{x} + c &=& 0\\\\ \dfrac{2}{x^2} + \dfrac{7}{x} + c &=& 0 \quad | \quad \cdot x^2 \\ 2 + 7x + cx^2 &=& 0 \\ cx^2+7x+2 &=& 0 \\\\ x &=& \dfrac{-7\pm\sqrt{7^2-4\cdot c\cdot 2}}{2c} \\ x &=& \dfrac{-7\pm\sqrt{49-4c}}{2c} \\ \dfrac{1}{r_1} + \dfrac{1}{r_2} &=& \dfrac{-7+\sqrt{49-4c}}{2c} + \dfrac{-7-\sqrt{49-4c}}{2c} \\\\ &=& \dfrac{-7+\sqrt{49-4c}-7-\sqrt{49-4c}}{2c} \\\\ &=& \dfrac{-14}{2c} \\\\ \mathbf{\dfrac{1}{r_1} + \dfrac{1}{r_2} } &\mathbf{=}& \mathbf{ -\dfrac{7}{c} } \\ \hline \end{array}$$

Dec 10, 2018
edited by heureka  Dec 11, 2018