Question

Find the quadratic polynomial whose zeroes are 2/3 and -1/4. Verify the relation between the

coefficients and the zeroes of the polynomial.​

Answer 1

Answer:

12x²-5x-2=0

Step-by-step explanation:

Let $\alpha$ = (2/3) and $\beta$ = (-1/4)

Then by formula for quadratic equation:

x² - (Sum of roots) + Products of roots = 0

x² - ($\alpha +\beta$) + ($\alpha \beta$) = 0   -------- (1)

Here, $\alpha + \beta$ = (2/3) + (-1/4) = (5/12)

and $\alpha \beta$ =(2/3)*(-1/4) = (-1/6)

Putting these value in equation (1), we get

x² - (5/12) + (-1/6) = 0

12x²-5x-2 = 0

Answer 2

Given:-

➜α=$\frac{2}{3}$

➜β=$\frac{-1}{4}$

$\rule{200}{1}$

To Find:-

☞The Quadratic Polynomial, and verify the relationship b/w the coefficients and its zeros?

$\rule{200}{1}$

AnsWer:-

✪Using Quadratic Formula✪

☞k[x²-(α+β)x+αβ]

↝α+β=$\frac{2}{3}$+($\frac{-1}{4}$)

★Taking LCM★

↝α+β=$\frac{2 \times 4}{3 \times 4}$+($\frac{-1 \times 3}{4 \times 3}$)

↝α+β=$\frac{8+(-3)}{12}$

↝α+β=$\frac{8-3}{12}$

☞α+β=$\frac{5}{12}$

$\rule{200}{1}$

↝αβ=$\frac{2}{3}$×($\frac{-1}{4}$)

↝αβ=$\frac{\cancel{-2}}{\cancel{12}}$

☞αβ=$\frac{-1}{6}$

$\rule{200}{1}$

▼Using The Values in the Formula▼

➜k[x²-($\frac{5}{12}$)x+($\frac{-1}{6}$)]

➜k[x²-$\frac{5}{12}$x-$\frac{1}{6}$ ]

♦Let k=12♦

➜12[x²-$\frac{5}{12}$x-$\frac{1}{6}$ ]

➜12x²-5x-2

☞12x²-5x-2 is the Required Polynomial.

$\rule{200}{2}$