For given roots of equation x^2-px+q=0, the condition q(3p+1)-q^2=0 is proved.
<h3>What is the meaning of the term root of the equation?</h3>
- The values of x that satisfy the following quadratic equation ax2 + bx + c = 0 are known as its roots.
- They are, in other words, the values of the parameter (x) that satisfy the equation.
- The roots of such a quadratic function are indeed the x-coordinates of the function's x-intercepts.
- Because the degree of such a quadratic equation is two, it can only have two roots.
For the given equation;
x^2-px+q=0
Let one roots of the equation be 'α'.
Then, other roots is the square of first; 'α²'.
The relation sum between the roots is;
α + α² = -coefficient of x
α + α² =-(-p)
α + α² = p .....eq1
And, the product of roots is,
α×α² = constant
α³ = q .....eq 2
From eq 1.
α + α² = p
Then, for the proof; p^3-q(3p+1)-q^2=0
α³ (α³ - 1 - 3(α² + α)) = 0
p^3 - q(3p+1) - q^2 = 0
Thus, for given roots of equation x^2-px+q=0, the condition q(3p+1)-q^2=0 is proved.
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