Question

Find the zeroes of 4x2+5√2x-3
by factorisation method and verify the relations between the zeroes and the coefficients of the polynomials.

Answer 1

hey friend here is your solution,

$4 {x}^{2} + 5 \sqrt{2} x - 3$
$4 {x}^{2} + 6\sqrt{2} x - \sqrt{2} x - 3$
$2 \sqrt{2} x( \sqrt{2} x + 3) - 1( \sqrt{2} x + 3)$

$(2 \sqrt{2} x - 1) \: \: ( \sqrt{2} x + 3)$

$2 \sqrt{2} x = 1 \: \: \: and \: \: \sqrt{2} x = - 3$

$x = \frac{1}{2 \sqrt{2}} \: \: and \: \: x = \frac{ - 3}{ \sqrt{2} }$

so these are the roots and the relationship is,

product of the zeroes,

$\alpha \times \beta = \frac{ c}{a}$

so,

$\frac{1}{2 \sqrt{2} } \times \frac{ - 3}{ \sqrt{2} } = \frac{ - 3}{4}$

and

sum of the zeroes,

$\alpha + \beta = \frac{ - b}{a}$
so,
$\frac{1}{2 \sqrt{2} } + \frac{ - 3}{ \sqrt{2} } = \frac{ - (\sqrt{2 } + ( - 3) 2\sqrt{2}) }{(2 \sqrt{2} )( \sqrt{2)} }$

then,
$\frac{ - ( \sqrt{2 } )+ ( - 6 \sqrt{2}) }{4}$

then,

$\frac{ - (\sqrt{2 } - 6 \sqrt{2} )}{4} = \frac{5 \sqrt{2} }{4}$


plz mark as brainliest.......

Answer 2

Hiii friend here is your answer,

P(X) = 4X²+5✓2X-3

=> 4X²+6✓2X-✓2X-3

=> 2✓2X(✓2X+3) -1(✓2X+3)

=> (✓2X+3) (2✓2X-1) = 0

=> (✓2X+3) = 0 OR (2✓2X-1) = 0

=> X = -3/✓2 OR X = 1/2✓2

-3/✓2 and 1/2✓2 are the two zeros of the given polynomial.

Let Alpha = -3/✓2 and beta = 1/2✓2

Relationship between the zeros and Coefficient.

Sum of Zeros= (Alpha + Beta) = -3/✓2 + 1/2✓2 = -3×2✓2 + ✓2 = -6✓2+✓2/4 = -5✓2/4 = -( Coefficient of X/Coefficient of X².

And,

Product of zeros = (-3/✓2 × 1/2✓2) = -3/4 = Constant term/Coefficient of X².

HOPE IT WILL HELP YOU...... :-)

Please mark it as brainliest!!