Question

N A quadratic polynomial, whose zeroes are - 3 and 4.is (A) r-x+ 12 (B) + x + 12 R2-6 6 (C) (D) 2r + 2x - 24 2 2 OA B oc OD​

Answer 1

EXPLANATION.

Quadratic polynomial.

Whose zeroes are = - 3 and 4.

As we know that,

Sum of the zeroes of the quadratic polynomial.

⇒ α + β = - b/a.

⇒ - 3 + 4 = 1.

⇒ α + β = 1.

Products of the zeroes of the quadratic polynomial.

⇒ αβ = c/a.

⇒ (-3) x (4) = - 12.

⇒ αβ = - 12.

As we know that,

Formula of quadratic polynomial.

⇒ x² - (α + β)x + αβ.

Put the values in the equation, we get.

⇒ x² - (1)x + (-12).

⇒ x² - x - 12.

                                                                                                                 

MORE INFORMATION.

Conjugate roots.

(1) = If D < 0.

One roots = α + iβ.

Other roots = α - iβ.

(2) = If D > 0.

One roots = α + √β.

Other roots = α - √β.

Answer 2

Answer:

Given ;-

• A quadratic polynomial whose zeroes are- 3 and 4.

To find: -

• A quadratic polynomial.

Solution :-

• Here one thing we know that,

• Sum of zeroes of quadratic polynomial ,

$\alpha + \beta = \frac{ - b}{ \: \: a}$

• Now applying all the values we get,

• $- 3 + 4 = 1$
• $\alpha + \beta = 1$

And we know that ,

• Product of zeroes Of quadratic polynomial is

$\alpha \beta = \frac{c}{a}$

• Now applying all the values we get,

• $- 3 \times 4 = - 12$
• $\alpha \beta = - 12$

Here,

• We know the quadratic equation formula that is ,

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

• Now taking the values we get that,

• ${x}^{2} - 1x + ( - 12) = { {x}^{2} - x - 12}^{2}$

Hope it helps u mate .

Thank you .