Question

Find the zeroes of the following following polynomials : 6x^2-3- 7x, 4u^2+8u^2, t^2-15

Answer 1

Answer:

for:6x^2-3-7x=0

x=3/2,

x=(-1/3)

Step by step explanation:-

6x^2-7x-3=0

6x^2+2x-9x-3=0

2x(3x+1)-3(3x+1)=0

(2x-3)(3x+1)=0

x=3/2

x=(-1/3)

Answer:

for:4u^2+8u^2

u=0

Step by step explanation:-

4u^2+8u^2=0

12u^2=0

u^2=0/12

u^2=0

u=0

Answer:

for:t^2-15

t=

$\sqrt{15}$

t=

$- \sqrt{15}$

Step by step explanation:-

t=

$(- b + - \sqrt{b^{2} - 4ac } ) \div 2a$

=

$(0 + - \sqrt{0 - 4 \times 1 \times ( - 15)} ) \div 2$

=

$(0 + - \sqrt{60} ) \div 2$

=

$+ - (2 \sqrt{15} ) \div 2$

=

$+ - \sqrt{15}$

that's why:

t=

$+ \sqrt{15}$

t=

$- \sqrt{15}$