Question

Find the zeroes of the quadratic polynomial x^{2} -3x-10 and verify the relationship between the zeroes and coefficient.

Answer 1

$\huge{\bf{\underline{\overline{\mathfrak{\red{Solution:-}\mid}}}}}$

$\tt{x}^{2} - 3x - 10$

$\tt = {x}^{2} - (5 - 2) x - 10$

$\tt = {x}^{2} - 5x + 2x - 10$

$\tt = x(x - 5) + 2(x - 5)$

$\tt= (x + 2)( x - 5)$

Either,

$x + 2 = 0 \: \: \: \: \: \: \: \: \: \: \: or \: \: \: \: \: \: \: \: \: x - 5 = 0 \\ x = - 2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = 5$

So, the reqd. zeros are :- 5 and -2

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Again,

• coefficient of x = -3
• coefficient of $\bf{x}^{2}$ = 1
• constant term = -10

Now,

$\boxed{Sum\:of\:zeros}$

$\tt = 5 + ( - 2)$

$\tt= 5 - 2$

$\tt= 3$

$\tt = \frac{ - ( - 3)}{ 1}$

$\bf\tt =\frac{ - \: coefficient \: of \: x}{coefficient \: o f\: {x}^{2} }$

And,

$\boxed{Product\:of\:zeros}$

$\tt = 5 \times ( - 2)$

$\tt= - 10$

$\tt= \frac{ -10 }{1}$

$\bf\tt= \frac{constant \: term}{coefficient \: of \: {x}^{2} }$

Answer 2

Step-by-step explanation:

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