Expand everything?

Algebra Level 3

a x 2 + b x + c = 0 \large ax^2+bx+c=0

a ( − b + b 2 − 4 a c 2 a ) 2 + b ( − b + b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a ) ( − b − b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a + − b − b 2 − 4 a c 2 a ) = ? \frac{a\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2+b\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\right) } = \ ?

a 2 a^2 b 2 b^2 a 2 b \dfrac{a^2}{b} b 2 c \dfrac{b^2}{c} c 2 a \dfrac{c^2}{a} c 2 c^2

Tommy Li by Tommy Li , 22, Hong Kong

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Viki Zeta
Sep 16, 2016

a x 2 + b x + c = 0 a x 2 + b x = − c .. (i) ⟹ x = α = − b + b 2 − 4 a c 2 a , x = β = − b − b 2 − 4 a c 2 a a ( x ) 2 + b ( x ) ( α ) ( β ) ( α + β ) = − c c a − b a = a 2 b ax^2 + bx + c = 0 \\ ax^2 + bx = -c \text{ .. (i) } \\ \implies x = \alpha = \dfrac{-b + \sqrt[]{b^2 - 4ac}}{2a}, x = \beta = \dfrac{-b - \sqrt[]{b^2-4ac}}{2a} \\ \dfrac{a(x)^2+b(x)}{(\alpha)(\beta)(\alpha+\beta) } \\ = \dfrac{-c}{\dfrac{c}{a} \dfrac{-b}{a}} \\ = \dfrac{a^2}{b}

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
Sep 16, 2016

We note that − b + b 2 − 4 a c 2 a \dfrac {-b + \sqrt{b^2-4ac}}{2a} and − b − b 2 − 4 a c 2 a \dfrac {-b - \sqrt{b^2-4ac}}{2a} are the two roots of a x 2 + b x + c = 0 ax^2+bx+c=0 . Let them be x 1 x_1 and x 2 x_2 respectively. Then, we have:

X = a ( − b + b 2 − 4 a c 2 a ) 2 + b ( − b + b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a ) ( − b − b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a + − b − b 2 − 4 a c 2 a ) = a x 1 2 + b x 1 x 1 x 2 ( x 1 + x 2 ) By Vieta’s formula x 1 x 2 = c a , x 1 + x 2 = − b a = \require c a n c e l a x 1 2 + b x 1 + c 0 − c c a ( − b a ) = a 2 b \begin{aligned} X & = \frac{a\left(\frac {-b + \sqrt{b^2-4ac}}{2a}\right)^2 + b\left(\frac {-b + \sqrt{b^2-4ac}}{2a}\right)}{\left(\frac {-b + \sqrt{b^2-4ac}}{2a}\right)\left(\frac {-b - \sqrt{b^2-4ac}}{2a}\right)\left(\frac {-b + \sqrt{b^2-4ac}}{2a} + \frac {-b - \sqrt{b^2-4ac}}{2a}\right)} \\ & = \frac {ax_1^2 + bx_1}{x_1x_2(x_1+x_2)} & \small \color{#3D99F6}{\text{By Vieta's formula }x_1x_2 = \frac ca, \ x_1+x_2 = - \frac ba} \\ & = \require {cancel} \frac {\cancel{ax_1^2 + bx_1+c}^0 - c}{\frac ca \left(-\frac ba \right)} \\ & = \boxed{\dfrac {a^2}b} \end{aligned}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

⟹ a x 2 + b x + c = 0 \implies \color{#EC7300}{ax^2+bx}+c=0

Let it roots be α \alpha and β \beta .

Using Quadratic formula .

⟹ x = α = − b + b 2 − 4 a c 2 a \implies x=\alpha=\dfrac{-b+\sqrt{b^2-4ac}}{2a}

And,

⟹ x = β = − b − b 2 − 4 a c 2 a \implies x=\beta=\dfrac{-b-\sqrt{b^2-4ac}}{2a}

Using Vieta's formula .

• α + β = − b a \ • \color{#3D99F6}{\alpha+\beta}=\dfrac{-b}{a}

• α β = c a \ • \color{#20A900}{ \alpha\beta}=\dfrac{c}{a}

We have to find:

⟹ a ( − b + b 2 − 4 a c 2 a ) 2 + b ( − b + b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a ) ( − b − b 2 − 4 a c 2 a ) ( − b + b 2 − 4 a c 2 a + − b − b 2 − 4 a c 2 a ) \implies \dfrac{a\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2+b\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)}{\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\left(\frac{-b+\sqrt{b^2-4ac}}{2a}+\frac{-b-\sqrt{b^2-4ac}}{2a}\right) }

= a x 2 + b x ( α β ) ( α + β ) =\dfrac{\color{#EC7300}{ax^2+bx}}{(\color{#20A900}{\alpha \beta})(\color{#3D99F6}{\alpha+\beta})}

⟹ − c c a × − b a = − a 2 c − b c = a 2 b \implies \cfrac{-c}{\dfrac{c}{a}×\dfrac{-b}{a}}=\dfrac{-a^2c}{-bc}=\boxed{\dfrac{a^2}{b}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...