Question

Find the zeros of the quadratic polynomial 5r^2 + 12t + 7 and verify the relationship between the zeros and the coefficients.

Answer 1

Answer:

Step-by-step explanation:

5t²+12t+7

=5t²+5t+7t+7 (splitting the middle term)

=5t(t+1)+7(t+1)

=(t+1)(5t+7)

t+1=0

Or t=-1

5t+7=0

Or t=-7/5

So the zeroes are -1 & -7/5

Sum of the zeroes=-1-7/5=-12/5 =-b/a

Product of the zeroes=-1×(-7/5)=7/5=c/a

Answer 2

Answer

The zeroes of the given polynomial are -7/5 and -1

Given

The quadratic polynomial is :

• 5t² + 12t + 7

Solution

Factorizing the given polynomial we have :

$\tt 5 {t}^{2} + 12t + 7 \\ \\ \longrightarrow \sf5 {t}^{2} + 5t + 7t + 7 \\ \\ \longrightarrow \sf \: 5t(t + 1) + 7(t + 1) \\ \\ \longrightarrow \sf (5t + 7)(t + 1)$

Therefore ,

$\implies \sf 5t + 7 = 0 \: \: \: and \implies t + 1 = 0 \\ \\ \sf\implies t = - \frac{7}{5} \: \: \: and \implies t = - 1$

Thus , the zeroes of the given polynomial are -7/5 and -1 respectively.

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Verification of relationships :

For the relation of sum of zeroes and the coefficients

$\sf sum \: of \: the \: zeroes = - \dfrac{coefficient \: of \: x}{coefficient \: of \: {x}^{2} }$

$\tt \implies - \dfrac{7}{5} - 1 = - \dfrac{ 12}{5} \\ \\ \tt\implies \dfrac{ - 7 - 5}{5} = - \dfrac{12}{5} \\ \\ \tt \implies - \dfrac{12}{5} = - \dfrac{12}{5}$

For the product of zeroes and the coefficients

$\sf product \: of \: the \: zeroes = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }$

$\sf\implies(- \dfrac{7}{5})\times (-1) = \dfrac{-7}{-5} \\\\ \sf\implies \dfrac{7}{5} = \dfrac{7}{5}$

$\bf\underline{Verified}$