Given that x2+ax+b&gt;0 and x2+(a+np)x+(b+nq)&gt;0 prove that x2+(a+∓)x+(b+mq)&gt;0,m=1,n−1―

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Given that x2+ax+b&amp;gt;0 and x2+(a+np)x+(b+nq)&amp;gt;0 prove that x2+(a+∓)x+(b+mq)&amp;gt;0,m=1,n−1―

Jared Kemp · Accepted Answer

Step 1Multiply the first inequality by&amp;nbsp;(n−m)&amp;nbsp;and the second by m and add these together(n−m)(x2+ax+b)&amp;gt;0m(x2+(a+np)x+(b+np))&amp;gt;0n(x2+(a+∓)x+(b+∓))&amp;gt;0Now divide the final inequality by&amp;nbsp;n.

Kamora Campbell · Accepted Answer

Step 1U could always check out&amp;nbsp;D.See if leading coefficient&amp;nbsp;&amp;gt;0&amp;nbsp;and&amp;nbsp;D&amp;nbsp;is less than 0 the quadratic is&amp;nbsp;&amp;gt;0&amp;nbsp;for all real&amp;nbsp;x.Given:&amp;nbsp;a2−4b&amp;lt;0Given:&amp;nbsp;(a+np)2-4(b+nq)=a2-4b+n(np2+2(ap-2q))&amp;lt;0(as&amp;nbsp;n&amp;nbsp;decreases&amp;nbsp;np2+2(ap−2q)&amp;nbsp;also decreases)&amp;nbsp;⇒&amp;nbsp;the expression is lesser for lesser &#039;n&#039; and hence less than 0.

Given that \(\displaystyle{x}^{{2}}+{a}{x}+{b}{>}{0}\) and \(\displaystyle{x}^{{2}}+{\left({a}+{n}{p}\right)}{x}+{\left({b}+{n}{q}\right)}{>}{0}\) prove

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