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JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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If x2+px+q=0{x^2} + px + q = 0 has the roots Ξ±β€…β€Šandβ€…β€ŠΞ²\alpha \;\mathrm{and}\;\beta, then the value of (Ξ±βˆ’Ξ²)2{\left( {\alpha - \beta } \right)^2} is equal to
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Correct option: A
p2βˆ’4q{p^2} - 4q
Option A (p2βˆ’4q{p^2} - 4q) is correct because it is the simplified form of the expression (Ξ±βˆ’Ξ²)2(\alpha - \beta)^2 using the relationship between roots and coefficients of a quadratic equation.

More Questions from QUADRATIC EQUATIONS

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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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