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26 Questions around this concept.
What is the value of $m$ so that both roots of the equation $x^2+m x+1$ are less than unity?
Let $\mathrm{f}(\mathrm{x})=a x^2+b x+c$, consider the following graph
Let $a_1<a_2<a_3<a_4$ then the number of real roots of equation $\left(x-a_1\right)\left(x-a_3\right)+\left(x-a_2\right)\left(x-a_4\right)=0$ equals
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Real values of m for which at least one root of $x^2-(m-3) x+m=0$ lies in $(0,1)$ (Given that 0 and 1 are not its roots)
Let f(x) = ax2 + bx + c , where a,b,c are real numbers and ‘a’ is non-zero number. Let ? and ? be the roots of the equation, and let k be a real number. Then:
1. If both roots of f(x) are less than k then
i) D ≥ 0 (as the real roots may be distinct or equal)
ii) af(k) > 0 (In both cases af(k) is positive, as in the second case if a < 0 then f(k) < 0, so multiplying two -ve values will give us a positive value)
iii)$\mathrm{k}>\frac{-\mathrm{b}}{2 \mathrm{a}}$ since, $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will lie between ? and ?, and ?, ? are less than k so $\frac{-b}{2 a}$ will be less than k.
2. If both roots of f(x) are greater than k
i) D ≥ 0 (as the real roots may be distinct or equal)
ii) af(k) > 0 (In both cases af(k) is positive, as in the second case if a < 0 then f(k) < 0, so multiplying two -ve values will give us a positive value)
iii)$\mathrm{k}<\frac{-\mathrm{b}}{2 \mathrm{a}}$ since $\frac{-\mathrm{b}}{2 \mathrm{a}}$ will lie between ? and $\underline{\underline{\underline{?}}}$, and $\underline{\underline{\underline{?}}}$ ? are greater than k so $\frac{-b}{2 a}$ will be greater than k .
Condition for number k
Let f(x) = ax2 + bx + c where a, b, c are real numbers and ‘a’ is a non-zero number. Let ? and ? be the real roots of the function. And let k be any real number. Then:
If k lies between the root ? and ?
af(k) < 0.
(As if a < 0 then f(k) > 0. So multiplying one -ve and one +ve value will give us a negative value)
Condition for numbers k1 and k2
Let f(x) = ax2 + bx + c, where a,b,c are real numbers and ‘a’ isa non-zero number. Let ? and ? be the real roots of the function. And let $k_1, k_2$ be any two real numbers. Then:
2. If exactly one root of f(x) lies in between the number $k_1, k_2$
$f\left(k_1\right) f\left(k_2\right)<0$ as for one value of k, we will have +ve value of f(x) and for other value of k, we will have a -ve value of f(x) (here ? < ?)
Condition on number $k_1, k_2$
Let f(x) = ax2 + bx + c, where a, b, c are real numbers and ‘a’ is non-zero number. Let ? and ? be the roots of the function. And let $k_1, k_2$ be any two real numbers. Then
1. If both roots lie between $k_1, k_2$
i) D ≥ 0 (as the real roots may be distinct or equal)
ii) $\begin{aligned} & \operatorname{af}\left(\mathrm{k}_1\right)>0 \text { and } \mathrm{af}\left(\mathrm{k}_2\right)>0 \\ & \mathrm{k}_1<\frac{-\mathrm{b}}{2 \mathrm{a}}<\mathrm{k}_2, \text { where } \alpha \leq \beta \text { and } \mathrm{k}_1<\mathrm{k}_2\end{aligned}$
2. If $k_1, k_2$ lies between the roots
$\mathrm{af}\left(\mathrm{k}_1\right)<0$ and $\mathrm{af}\left(\mathrm{k}_2\right)<0$
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