We need to make h alone.
Add 7 to both sides to remove -7 from the right, and multiply both sides by -1 to make h positive. Doesn't matter which you do first.
Result:
6=h
h=6

6 0

3 years ago

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What are the roots to<br> the equation<br> 2x² + 6x-1=0?

lys-0071 [83]

Answer:

The roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = $\frac{ - b}{a}$

And products of roots = $\frac{c}{a}$

So, m + n = $\frac{ - 6}{2}$ = - 3

And m × n = $\frac{ - 1}{2}$

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 ( $\frac{ - 1}{2}$ )

Or, (m - n)² = 9 + 2 = 11

I.e m - n = $\sqrt{11}$

Again m + n = - 3 And m - n = $\sqrt{11}$

Solving this two equation

(m + n) + ( m - n) = - 3 + $\sqrt{11}$

I.e 2 m = - 3 + $\sqrt{11}$

Or, m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$

Similarly n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Hence the roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$ Answer

6 0

3 years ago