CONCEPT: If one CONJUGATE ROOT of a quadratic EQUATION is α = \(\sqrt{a} + b\)Then other roots will be, β = \(\sqrt{a}-b\) Then, we can WRITE the quadratic equation as, x2 - (α + β) x + α β = 0Calculation:Given one of the roots of quadratic having rational coefficients is α = √7 -4We know, the quadratic equations will have conjugate irrational roots. then second root will be β = √7 + 4 Then sum of roots are α + β = √7 - 4 + √7 + 4 = 2√7and product of roots α.β = (√7 - 4) (√7 + 4) = -9Then quadratic equation will be,⇒ x2 - (α + β) x + αβ = 0 ⇒ x2 - 2√7x - 9 = 0 "> CONCEPT: If one CONJUGATE ROOT of a quadratic EQUATION is α = \(\sqrt{a} + b\)Then other roots will be, β = \(\sqrt{a}-b\) Then, we can WRITE the quadratic equation as, x2 - (α + β) x + α β = 0Calculation:Given one of the roots of quadratic having rational coefficients is α = √7 -4We know, the quadratic equations will have conjugate irrational roots. then second root will be β = √7 + 4 Then sum of roots are α + β = √7 - 4 + √7 + 4 = 2√7and product of roots α.β = (√7 - 4) (√7 + 4) = -9Then quadratic equation will be,⇒ x2 - (α + β) x + αβ = 0 ⇒ x2 - 2√7x - 9 = 0 ">

If one of the roots of quadratic having rational coefficients is √7 - 4, then the quadratic equation is:

General Aptitude Algebra in General Aptitude 10 months ago

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CONCEPT: If one CONJUGATE ROOT of a quadratic EQUATION is α = \(\sqrt{a} + b\)Then other roots will be, β = \(\sqrt{a}-b\) Then, we can WRITE the quadratic equation as, x2 - (α + β) x + α β = 0Calculation:Given one of the roots of quadratic having rational coefficients is α = √7 -4We know, the quadratic equations will have conjugate irrational roots. then second root will be β = √7 + 4 Then sum of roots are α + β = √7 - 4 + √7 + 4 = 2√7and product of roots α.β = (√7 - 4) (√7 + 4) = -9Then quadratic equation will be,⇒ x2 - (α + β) x + αβ = 0 ⇒ x2 - 2√7x - 9 = 0

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