Question

One root of the equation x^2-px+q=0 is the square of the Other. Show that p^3-q(3p+1)-q^2=0

Answer 1

For given roots of equation x^2-px+q=0, the condition q(3p+1)-q^2=0 is proved.

What is the meaning of the term root of the equation?

• 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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