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JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1
If one of the roots of the equation x2+bx+3=0{x^2} + bx + 3 = 0 is thrice the other, then bb is equal to
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Correct option: D
Β±4\pm 4
Option D (Β±4\pm 4) is correct because by setting up the relationship between the roots and using Vieta's formulas, we can solve for the value of b.

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The solution of the equation log⁑7log⁑5(x+5+x)=0\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0 is If Ξ±,Ξ²\alpha ,\beta are the roots of x2βˆ’3x+a=0,a∈R{x^2} - 3x + a = 0,a \in R and Ξ±<1<Ξ²,\alpha < 1 < \beta , then If Ξ±,β€…β€ŠΞ²\alpha ,\;\beta be the roots of the equation x2βˆ’px+q=0{x^2} - px + q = 0 and Ξ±1,Ξ²1{\alpha _1},{\beta _1} be the roots of the equation x2βˆ’qx+p=0,{x^2} - qx + p = 0, then the equation whose roots are 1Ξ±1Ξ²+1Ξ±Ξ²1andβ€…β€Š1Ξ±Ξ±1+1Ξ²Ξ²1,\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}}, is If x2+4ax+2>0{x^2} + 4ax + 2 > 0 for all values of xx, then aa lies in the interval If f(x)=ax2+bx+cβ€…β€Šf\left( x \right) = a{x^2} + bx + c\; and β€…β€Šf(βˆ’1)<1,f(1)>βˆ’1,f(3)<βˆ’4β€…β€Š\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\; and β€…β€Šaβ‰ 0,\;a \ne 0, then Root(s) of the equation 9x2βˆ’18∣x∣+5=09{x^2} - 18\left| x \right| + 5 = 0 belonging to the domain of definition of the function f(x)=log⁑(x2βˆ’xβˆ’2)f\left( x \right) = \log \left( {{x^2} - x - 2} \right) is (are)
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