Juan Hewlett

2021-12-15

Find a quadratic polynomial whose sum and product respectively of the zeroes are as given also find the zeoes of these polynomial by factorization

(-8/3),(4/3)

(-8/3),(4/3)

Durst37

Beginner2021-12-16Added 37 answers

Step 1 Given

$\left(\frac{-8}{3}\right),\frac{4}{3}$

Step 2 Finding the zeroes

Sum of the zeroes$=\frac{-8}{3}$

Product of the zeroes$=\frac{4}{3}$

$p\left(x\right)={x}^{2}-(\sum \text{}of\text{}the\text{}zeroes)+(\prod uct\text{}of\text{}the\text{}zeroes)$

Then,

$p\left(x\right)={x}^{2}-\frac{8x}{3}+\frac{4}{3}$

$p\left(x\right)=3{x}^{2}-8x+4$

By factorization

$3{x}^{2}-8x+4=0$

$3{x}^{2}-(6x+2x)+4=0$

$3{x}^{2}-6x-2x+4=0$

3x(x-2)-2(x-2)=0

(x-2)(3x-2)=0

x-2=0; 3x-2=0

$x=2;x=\frac{2}{3}$

$\therefore x=2;\frac{2}{3}$

Step 2 Finding the zeroes

Sum of the zeroes

Product of the zeroes

Then,

By factorization

3x(x-2)-2(x-2)=0

(x-2)(3x-2)=0

x-2=0; 3x-2=0

Maricela Alarcon

Beginner2021-12-17Added 28 answers

Step 1

Let the roots of a quadratic equation

$a{x}^{2}+bx=c=0$ be $\alpha \text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}\beta$

So,$\alpha +\beta =\frac{-b}{a}=\frac{-8}{3}$ ...(1)

$\alpha \beta =\frac{c}{a}=\frac{4}{3}$ ...(2)

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

$a({x}^{2}+\frac{b}{a}x+\frac{c}{a})=0$

Step 2

${x}^{2}-\left(\frac{-b}{a}\right)x+\frac{c}{a}=0$

${x}^{2}-(\alpha +\beta )x+\alpha \beta =0$

Substitute (1) and (2) in above

${x}^{2}-\left(\frac{-8}{3}\right)x+\frac{4}{3}=0$

$3{x}^{2}+8x+4=0$

$\therefore$ The quadratic equation is $3{x}^{2}+8x+4=0$

Step 3

$3{x}^{2}+8x+4=0$

$3{x}^{2}+2x+6x+4=0$

x(3x+2)+2(3x+2)=0

(3x+2)(x+2)=0

$x=\frac{-2}{3},x=-2$

$\therefore$ The roots are $\frac{-2}{3},-2$

Let the roots of a quadratic equation

So,

Step 2

Substitute (1) and (2) in above

Step 3

x(3x+2)+2(3x+2)=0

(3x+2)(x+2)=0

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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