Question

H(t)= -16t^2+150+35 algebra question

Answer 1

t = sqrt(185)/4 or t = -sqrt(185)/4

Explanation:

Solve for t over the real numbers:

-16 t^2 + 35 + 150 = 0

-16 t^2 + 35 + 150 = -16 t^2 + 185:

-16 t^2 + 185 = 0

Multiply both sides by -1:

16 t^2 - 185 = 0

t = (0 ± sqrt(0^2 - 4×16 (-185)))/(2×16) = ( ± sqrt(11840))/32:

t = sqrt(11840)/32 or t = (-sqrt(11840))/32

sqrt(11840) = sqrt(64×5×37) = sqrt(2^6×5×37) = 2^3sqrt(5×37) = 8 sqrt(185):

t = (8 sqrt(185))/32 or t = (-8 sqrt(185))/32

(8 sqrt(185))/32 = (8 sqrt(185))/(8×4) = (sqrt(185))/4:

t = sqrt(185)/4 or t = (-8 sqrt(185))/32

(8 (-sqrt(185)))/32 = (8 (-sqrt(185)))/(8×4) = -(sqrt(185))/4:

Answer: t = sqrt(185)/4 or t = -sqrt(185)/4

Answer 2

`t=(75+√(6185))/(16),\quad t=(75-√(6185))/(16)`

Explanation:

Given quadratic function:

`h(t)=-16t^2+150t+35`

To solve the quadratic, set it to zero and solve for t using the quadratic formula.

`\boxed{\begin{minipage}{4 cm}\underline{Quadratic Formula}\\\\$x=(-b \pm √(b^2-4ac))/(2a)$\\\\when $ax^2+bx+c=0$ \\\end{minipage}}`

Compare the coefficients of the given quadratic function:

• a = -16
• b = 150
• c = 35

Substitute the values of a, b and c into the quadratic formula:

`t=(-150 \pm √(150^2-4(-16)(35)))/(2(-16))`

Simplify:

`t=(-150 \pm √(22500+2240))/(-32)`

`t=(150 \pm √(24740))/(32)`

Rewrite 24740 as 2² · 6185:

`t=(150 \pm √(2^2 \cdot 6185))/(32)`

`\textsf{Apply the radical rule:} \quad √(ab)=√(a)√(b)`

`t=(150 \pm √(2^2) √(6185))/(32)`

`\textsf{Apply the radical rule:} \quad √(a^2)=a, \quad a \geq 0`

`t=(150 \pm 2 √(6185))/(32)`

Divide all terms by 2:

`t=(75\pm √(6185))/(16)`

Therefore, the exact solutions to the quadratic equation are:

`\boxed{\boxed{t=(75+√(6185))/(16),\quad t=(75-√(6185))/(16)}}`