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Question: Under which one of the following conditions will the quadratic equation \({{x}^{2}}+mx+2=0\) always ...

Under which one of the following conditions will the quadratic equation x2+mx+2=0{{x}^{2}}+mx+2=0 always have real roots?
A. 23≤m2<82\sqrt{3}\le {{m}^{2}}<8
B. 3≤m2<4\sqrt{3}\le {{m}^{2}}<4
C. m2≥8{{m}^{2}}\ge 8
D. m2≤3{{m}^{2}}\le \sqrt{3}

Explanation

Solution

For solving this question you should know about the quadratic equations and to find the roots of them. This b2−4ac{{b}^{2}}-4ac is part of the quadratic formula, which is denoted at the place of discriminant. Find this discriminant here denoted as ‘D’. If we find the roots of any quadratic equation, then we use this formula. And if b2−4ac≥0{{b}^{2}}-4ac\ge 0, then the quadratic equation has real roots.

Complete step-by-step solution:
According to the question we have to find a condition for which our quadratic equation x2+mx+2=0{{x}^{2}}+mx+2=0 always has real roots. So, as we know, the quadratic equations of the format ax2+bx+ca{{x}^{2}}+bx+c always contain roots which can be imaginary or real roots. But in many equations, we can’t determine roots easily. So, then we use the formula for determining the roots. The quadratic formula is given as:
−b±b2−4ac2a\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}
But discriminant D=b2−4acD={{b}^{2}}-4ac
So, we can write the quadratic formula as:
−b±D2a\dfrac{-b\pm \sqrt{D}}{2a}
And here we get two roots as:
−b+b2−4ac2a\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a} and −b−b2−4ac2a\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}
If we take an example to understand it clearly then:
Example 1: Find the roots of x2−4x+6{{x}^{2}}-4x+6.
The discriminant is used to determine how many different solutions and what type of solutions a quadratic equation will have. So, here according to the question,
a=1,b=−4,c=6 ⇒D=(−4)2−4(1)(6) ⇒D=16−24 ⇒D=−8 \begin{aligned} & a=1,b=-4,c=6 \\\ & \Rightarrow D={{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 6 \right) \\\ & \Rightarrow D=16-24 \\\ & \Rightarrow D=-8 \\\ \end{aligned}
And if we have to find only the real roots of any quadratic equation, then the condition is,
b2−4ac≥0{{b}^{2}}-4ac\ge 0
If we apply this condition in our given equation, x2+mx+2=0{{x}^{2}}+mx+2=0, then, for real roots, b2−4ac≥0{{b}^{2}}-4ac\ge 0,
⇒m2−4(1)(2)≥0 ⇒m2−8≥0 ⇒m2≥8 \begin{aligned} & \Rightarrow {{m}^{2}}-4\left( 1 \right)\left( 2 \right)\ge 0 \\\ & \Rightarrow {{m}^{2}}-8\ge 0 \\\ & \Rightarrow {{m}^{2}}\ge 8 \\\ \end{aligned}
So, the correct option is C.

Note: If we want to calculate the roots of any equation then always try to reduce that in a form of ax2+bx+ca{{x}^{2}}+bx+c because it will be very easy to find the roots from this. And it gives us exact roots, so always follow this method.