Question

Find the quadratic polynomial whose zeroes are 7 and -3

a)x^2-4x-21
b)x^2+4x+21
c)x^2-4x+21
d)x^2-21x-4
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Answer 1

Step-by-step explanation:

x^2-49=0 I know the answer is 7,-7 but I am not able to get that. please show me the steps so I can see what I am missing thank you

x2 - 49 = 0

Easiest solution:

x2 = 49 [Add 49 to both sides]

√x2 = √49 [Take the square root of both sides]

x = ±7

Using quadratic formula:

-(b/2a)±[b2-4ac]1/2/2a [for ax2+bx+c=0]

Your equation is (1)x2 + (0)x + (-49) = 0

so a=1, b=0, c=-49. Plugging them into the quadratic formula:

-(0/2)±[02-4(1)(-49)]1/2/2

0 ± [4*49]1/2/2

± (2*7)/2

x = ±7

Answer 2

Given ,

The zeroes of quadratic polynomial are 7 and -3

We know that ,

The quadratic polynomial whose zeroes are given can be written as

$\large \sf \fbox{ {x}^{2} - (sum \: of \: roots)x + (product \: of \: roots)}$

Thus ,

x² - {7 + (-3)}x + {7 × (-3)}

x² - 4x - 21

$\sf \therefore \underline{ The \: quadratic \: polynomial \: is \: {x}^{2} - 4x - 21}$