Question

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients of 5x2 -4-8x

Answer 1

EXPLANATION.

Equation are = 5x² - 8x - 4 = 0

equation is in the form of = ax² + bx + c

evaluate into middle term split.

=> 5x² - 8x - 4 = 0

=> 5x² - 10x + 2x - 4 = 0

=> 5x ( x - 2 ) + 2 ( x - 2 ) = 0

=> ( 5x + 2 ) ( x - 2 ) = 0

=> x = -2/5 and x = 2

Let a = -2/5

Let b = 2

Now,

sum of zeroes of quadratic equation

=> a + b = -b/a

=> -2/5 + 2 = 8/5

=> 8/5 = 8/5

products of zeroes of quadratic equation

=> ab = c/a

=> -2/5 X 2 = -4/5

=> -4/5 = -4/5

HENCE VERIFIED.

Answer 2

Solution:-

Factories the polynomial

$\rm \: 5 {x}^{2} - 8x - 4 = 0$

By splitting into middle term , we get

$\rm \: 5 {x}^{2} - 10x + 2x - 4 = 0$

$\rm \: 5x(x - 2) + 2(x - 2) = 0$

$\rm \: (5x + 2)(x - 2) = 0$

Zeroes of polynomial are

First zeroes are

$\rm \: 5x + 2 = 0$

$\rm \: 5x = - 2$

$\rm \: x = \frac{ - 2}{5}$

Second zeroes are

$\rm \: x - 2 = 0$

$\rm \: x = 2$

So

$\rm \alpha = \frac{ - 2}{5} \: \: and \: \: \beta = 2$

Eq

$\rm \: 5 {x}^{2} - 8x - 4 = 0$

Compare with

$\rm \: a {x}^{2} + bx + c = 0$

Now

$\rm \: a = 5 \: \: b = - 8 \: \: \: and \: \: c = - 4$

Sum of the zeroes is

$\rm \: \alpha + \beta = \frac{ - coffecient \: of \: x}{coffecient \: \: of \: {x}^{2} }$

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

$\rm \: \frac{ - 2}{5} + 2 = \frac{ - ( - 8)}{5}$

Taking lcm of Lhs side

$\rm \: \frac{ - 2 + 10}{5} = \frac{8}{5}$

$\rm \: \frac{8}{5} = \frac{8}{5}$

LHS = RHS

Product of zero are

$\rm \: \alpha \times \beta = \frac{constant \: term}{cofficient \: of \: {x}^{2} }$

$\rm \: \frac{ - 2}{5} \times 2 = \frac{ - 4}{5}$

$\rm \: \frac{ - 4}{5} = \frac{ - 4}{5}$

LHS = RHS