Question

find the zeroes of the quadratic polynomial X2+7x-30 & varify the relationship between the zeroes & the coefficients
(please give me correct answer and fast I need help please)​

Answer 1

S O L U T I O N :

We have quadratic equation or polynomial p(x) = x² + 7x - 30 & zero of the polynomial p(x) = 0.

$\underline{\underline{\tt{By\:quadratic\:formula\::}}}$

$\boxed{\bf{x = \frac{-b\pm \sqrt{b^{2} - 4ac} }{2a}}}$

As we know that given equation compared with ax² + bx + c;

• a = 1
• b = 7
• c = -30

Now,

$\mapsto\tt{x = \dfrac{-b\pm\sqrt{b^{2} - 4ac} }{2a} }$

$\mapsto\tt{x = \dfrac{-7\pm\sqrt{(7)^{2} - 4\times 1 \times (-30)} }{2\times 1} }$

$\mapsto\tt{x = \dfrac{-7\pm\sqrt{49 +120} }{2} }$

$\mapsto\tt{x = \dfrac{-7\pm\sqrt{169} }{2} }$

$\mapsto\tt{x = \dfrac{-7\pm 13 }{2} }$

$\mapsto\tt{x = \dfrac{-7 + 13}{2} \:\:\:Or\:\:\: x = \dfrac{-7-13}{2} }$

$\mapsto\tt{x = \dfrac{6}{2} \:\:\:Or\:\:\: x = \dfrac{-20}{2} }$

$\mapsto\tt{x = \cancel{\dfrac{6}{2}} \:\:\:Or\:\:\: x = \cancel{\dfrac{-20}{2}} }$

$\mapsto\bf{x = 3\:\:Or\:\:x = -10}$

∴ α = 10 & β = -3 are two zeroes of the given polynomial .

V E R I F I C A T I O N :

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta = \dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{3+ (-10) = \dfrac{-7}{1}}$

$\mapsto\tt{3-10 = -7}$

$\mapsto\bf{-7 = -7}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{3\times (-10) = \dfrac{-30}{1}}$

$\mapsto\bf{-30 = -30}$

Thus,

The relationship between zeroes & coefficient are verified .

Answer 2

Answer:

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Step-by-step explanation:

here ur answer