given x1>0 and x2=4x1 are solutions of the quadratic equation ax^2+bx+c=0. if 3a =2(c-b) then the value of x1is equal to:
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ANSWER
Let us recall the general solution, α = (-b-√b2-4ac)/2a and β = (-b+√b2-4ac)/2a
Case I: D^2 = b2 – 4ac > 0
When a, b, and c are real numbers, a ≠ 0 and discriminant is positive, then the roots α and β of the quadratic equation ax2 +bx+ c = 0 are real and unequal.
Case II: b2– 4ac = 0
When a, b, and c are real numbers, a ≠ 0 and discriminant is zero, then the roots α and β of the quadratic equation ax2+ bx + c = 0 are real and equal.
Case III: b2– 4ac < 0
When a, b, and c are real numbers, a ≠ 0 and discriminant is negative, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are unequal and not real. In this case, we say that the roots are imaginary.
Case IV: b2 – 4ac > 0 and perfect square
When a, b, and c are real numbers, a ≠ 0 and discriminant is positive and perfect square, then the roots α and β of the quadratic equation ax2 + bx + c = 0 are real, rational and unequal.
Case V: b2– 4ac > 0 and not perfect square
When a, b, and c are real numbers, a ≠ 0 and discriminant is positive but not a perfect square then the roots of the quadratic equation ax2 + bx + c = 0 are real, irrational and unequal.
Here the roots α and β form a pair of irrational conjugates.
Case VI: b2– 4ac >0 is perfect square and a or b is irrational
When a, b, and c are real numbers, a ≠ 0 and the discriminant is a perfect square but any one of a or b is irrational then the roots of the quadratic equation ax2 + bx + c = 0 are irrational.