Question

If zeroes of quadratic Polynomial is -3, ¼ find quadratic Polynomial​

Answer 1

Answer:

$4 {x}^{2} - 11x - 3 = 0$

Step-by-step explanation:

$α=(-3), β= \frac{1}{4} \\ {x}^{2} +(α+β)x+αβ=0 \\ {x}^{2} +(-3+ \frac{1}{4} )x+(-3)( \frac{1}{4} )=0 \\ {x}^{2} + ( \frac{ - 12 + 1}{4} )x + ( \frac{ - 3}{4} ) = 0 \\ {x}^{2} + ( \frac{ - 11}{4} )x + ( \frac{ - 3}{4} ) = 0 \\ {x}^{2} - \frac{11x}{4} - \frac{3}{4} = 0 \\ \frac{4 {x}^{2} - 11x - 3 }{4} = 0 \\ 4 {x}^{2} - 11x - 3= 0 \\ Therefore, \: the \: quadratic \: polynomial \: is \: 4 {x}^{2} - 11x - 3 = 0.$

Answer 2

Answer:  4x²+12x+1

Step-by-step explanation: ( sum of zeroes )α+β = -3

                                           ( product of zeroes)α*β = 1/4

[x²-(sum of zeroes)x + product of zeroes]

x²-(-3)x + 1/4

x²+3x=1/4

4 [ x²+3x+1/4]

4x²+12x+ 4*1/4 ( 4 will be canceled because numerator and denominator are the same )

so, our quadratic equation will be 4x² + 12x + 1