Question

How do I solve quadratic equations for example,h(t) = – 0.07t2 + 0.007t + 5

Answer 1

`t=8.50, -8.40\:\: \sf (2 \:d.p.)`

Explanation:

Quadratic equations are in the form
`ax^2+bx+c=0`, where
`a\\eq 0`.

There are three basic methods to solve quadratic equations algebraically:

• Factoring
• Completing the Square
• Quadratic Formula

Given quadratic equation:

`h(t)=-0.07t^2+0.007t+5`

Method 1 - Factoring

If the trinomial is able to be easily factored, set the equation equal to zero and factor. Set each factor equal to zero and solve.

As the given example is not easily factored, factoring is not the best method to use on this occasion.

Method 2 - Completing the Square

If the quadratic expression is not able to be easily factored, try completing the square.

Divide all terms by
`a` (-0.07):

`\implies t^2-(1)/(10)t-(500)/(7)=0`

Move the constant to the right side:

`\implies t^2-(1)/(10)t=(500)/(7)`

Add the square of half the coefficient of t to both sides:

`\implies t^2-(1)/(10)t+\left(-(1)/(20)\right)^2=(500)/(7)+\left(-(1)/(20)\right)^2`

`\implies t^2-(1)/(10)t+(1)/(400)=(200007)/(2800)`

Factor the perfect trinomial:

`\implies \left(t-(1)/(20)\right)^2=(200007)/(2800)`

Square root both sides:

`\implies t-(1)/(20)=\pm\sqrt{(200007)/(2800)`

Add 1/20 to both sides:

`\implies t=(1)/(20)\pm\sqrt{(200007)/(2800)`

As decimals to 2 decimal places:

`\implies t=8.50, -8.40\:\:\sf(2 \:d.p.)`

Method 3 - Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)`

This can be used to find both real and imaginary or complex solutions.

Identity the variables:

`a=-0.07, \quad b=0.007, \quad c=5`

Substitute the values into the quadratic formula and solve for t:

`\implies t=(-0.007 \pm √(0.007^2-4(-0.07)(5)) )/(2(-0.07))`

`\implies t=(-0.007 \pm √(1.400049))/(-0.14)`

`\implies t=(0.007 \pm √(1.400049))/(0.14)`

`\implies t=8.50, -8.40\:\: \sf (2 \:d.p.)`

Finally, a non-algebraic method of solving a quadratic equation is by graphing using a graphical calculator (see attached). The solutions are the points at which the curve intercepts the x-axis.

Answer 2

x ≈ -8.402, 8.502

Explanation:

The usual methods of solving quadratic equations apply to this one. Any of them can be used: graphing, factoring, completing the square, quadratic formula.

Solution

Often, when the coefficients are non-integers and have no obvious relationship to each other, the most convenient method of solution is the quadratic formula. It tells you the solution to ax²+bx+c=0 is ...

`x=(-b\pm√(b^2-4ac))/(2a)`

Your equation has the coefficients, a = -0.07, b = 0.007, c = 5, so the solutions are ...

`x=(-0.007\pm√(0.007^2-4(-0.07)(5)))/(2(-0.07))=(-0.007\pm√(1.400049))/(-0.14)\\\\x=0.05\pm\sqrt{(200007)/(2800)}=0.05\pm\sqrt{71.4310\overline{714285}}\\\\\boxed{x\approx\{-8.40169,\,8.50169\}}`