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# Help pls

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Let $p$ and $q$ be real numbers such that the roots of $x^2 + px + q = 0$are real. Prove that the roots of $x^2 + px + q + (x + a)(2x + p) = 0$are real, for any real number $a.$

Jun 26, 2020

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Is the second quadratic $$x^2+px+q+(x+a)(2x+p)=0$$? Or something else? please clarify.

Jun 27, 2020
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$$x^2+px+q=0\qquad \text{has real roots}\\so\\ p^2-4q\ge0\\ p^2-4q\ge0\\ q\le\frac{p^2}{4}\\ q=\frac{p^2}{4}-\delta\qquad \text{Where }\delta\in R\;\;\; and \;\;\; \delta \ge0\\$$

Prove that the following quadratic must have real roots.

$$x^2+px+q+(x+a)(2x+p)=0\\ \text{I simplified this to}\\ 3x^2+(2a+2p)x+q+ap=0\\ \text{This will have real roots if the discriminant is greater or equal to 0}\\ \triangle=(2a+2p)^2-12(q+ap)\\ ...\\ \triangle=4[a^2+p^2-ap-3q]\\ \triangle=4[a^2+p^2-ap-3(\frac{p^2}{4}-\delta)]\\ \triangle=4[a^2+\frac{p^2}{4}-ap+3\delta]\\ \triangle=4[(a-\frac{p}{2})^2+3\delta]\\ \triangle \ge0\\ \therefore \text{the second quadratic has real roots also}\\ QED$$

Note: I have not checked this proof properly, it could have errors.

LaTex:

x^2+px+q=0\qquad \text{has real roots}\\so\\
p^2-4q\ge0\\
p^2-4q\ge0\\
q\le\frac{p^2}{4}\\
q=\frac{p^2}{4}-\delta\qquad \text{Where }\delta\in R\;\;\; and \;\;\; \delta \ge0\\

x^2+px+q+(x+a)(2x+p)=0\\ \text{I simplified this to}\\ 3x^2+(2a+2p)x+q+ap=0\\ \text{This will have real roots if the discriminant is greater or equal to 0}\\ \triangle=(2a+2p)^2-12(q+ap)\\ ...\\ \triangle=4[a^2+p^2-ap-3q]\\ \triangle=4[a^2+p^2-ap-3(\frac{p^2}{4}-\delta)]\\ \triangle=4[a^2+\frac{p^2}{4}-ap+3\delta]\\ \triangle=4[(a-\frac{p}{2})^2+3\delta]\\ \triangle \ge0\\ \therefore \text{the second quadratic has real roots also}\\ QED

Jun 27, 2020