Proving Quadratic Equation

**saberteeth** · August 28th 2009, 03:35 AM

How do i prove this?

If one root of the equation ax^2 + bx + c = 0 is square of the other, show that b^3 + a^2.c + a.c^2 = 3abc

**Taluivren** · August 28th 2009, 04:03 AM

If x,y are roots and one is the square of the other, then $x-y^2 =0$ or $x^2 - y = 0$ . It follows that $0=(x-y^2)(x^2-y)= x^3 +y^3 -xy -x^2y^2 = (x+y)^3 - 3xy^2-3x^2y -xy -x^2y^2 =$
$<br /> = (x+y)^3 -3xy(x+y) -xy -x^2y^2 <br />$
Now Viete's formulas say $x+y = -\frac{b}{a}$ and $xy=\frac{c}{a}$ .
So $0=-\frac{b^3}{a^3} -3\frac{c}{a}(-\frac{b}{a}) - \frac{c}{a} - \frac{c^2}{a^2} = -\frac{b^3}{a^3} +3\frac{cb}{a^2}- \frac{c}{a} - \frac{c^2}{a^2}$ from which by multipliyng by $a^3$ the conclusion follows.

**DeMath** · August 28th 2009, 04:33 AM

Originally Posted by saberteeth

How do i prove this?

If one root of the equation ax^2 + bx + c = 0 is square of the other, show that b^3 + a^2.c + a.c^2 = 3abc

Or

$a{x^2} + bx + c = 0.$

${x_1} = x_2^2 \Leftrightarrow \frac{{ - b + \sqrt {{b^2} - 4ac} }}<br /> {{2a}} = {\left( {\frac{{ - b - \sqrt {{b^2} - 4ac} }}<br /> {{2a}}} \right)^2} \Rightarrow$

$\Rightarrow - ab + a\sqrt {{b^2} - 4ac} = {b^2} + b\sqrt {{b^2} - 4ac} - 2ac \Leftrightarrow$

$\Leftrightarrow \left( {a - b} \right)\sqrt {{b^2} - 4ac} = {b^2} + ab - 2ac \Rightarrow$

$\Rightarrow \left( {{a^2} - 2ab + {b^2}} \right)\left( {{b^2} - 4ac} \right) = {b^4} + 2{b^2}\left( {ab - 2ac} \right) + {\left( {ab - 2ac} \right)^2} \Leftrightarrow$

$\Leftrightarrow {a^2}{b^2} - 4{a^3}c - 2a{b^3} + 8{a^2}bc + {b^4} - 4{b^2}ac =$

$= {b^4} + 2a{b^3} - 4a{b^2}c + {a^2}{b^2} - 4{a^2}bc + 4{a^2}{c^2} \Leftrightarrow$

$- 4{a^3}c - 2a{b^3} + 8{a^2}bc = 2a{b^3} - 4{a^2}bc + 4{a^2}{c^2} \Leftrightarrow$

$4a{b^3} + 4{a^2}{c^2} + 4{a^3}c = 12{a^2}bc \Leftrightarrow$

$\Leftrightarrow {b^3} + {a^2}c + a{c^2} = 3abc.$

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