Homework Clinic
Mathematics Clinic => Grade 9 Mathematics => Topic started by: erika on Oct 3, 2013
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How do you solve the equation using the quadratic formula?
3x^2 -17=0
or
2y^2=-5y
Thanks for your help !!
Thanks for all the answers. It makes perfect sense now.
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2y^2+5y=0
y(2y+5=0
y=0 or -5/2
3x^2-17=0
3x^2=17
x^2=17/3
x= sq rt 17/3
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x= [ -b +/- sqr(b^2 - 4ac) ] / 2a
where:
ax^2 + bx + c = 0
3x^2 -17=0
a= 3
b= 0
c= -17
plug and chug:
x= 2.3805, -2.3805
2y^2=-5y
2y^2 - 5y = 0
a= 2
b= -5
c= 0
plug and chug:
y= 2.5, 0
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The quadratic formula says:
[-b ± ?(b²-4ab)] / 2a
1)
3x² - 17 = 0
So in this case:
a = 3
b = 0
c = -17
Plug them into the quadratic equation:
[0 ± ?(0 - 4(3)(-17))] / 6
[±?(204)] / 6
[±2?51] / 6
[±?51] / 3 <===Answer
2)
First rearrange the equation:
2y² + 5y = 0
So in this case:
a = 2
b = 5
c = 0
Plug it into the quadratic equation:
[-5 ± ?(25 - 4(2)(0))] / 4
[-5 ± ?25] / 4
[-5 ± 5] / 4
So you have to equations to solve for. I'll do the plus one first:
[-5 + 5] / 4=
0 <===Answer #1
Now the minus one:
[-5 - 5] / 4=
-5/2 <===Answer #2
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for 3x^2 - 17 = 0
ax^2 + bx + c = 0
a = 3
b = 0
c = -17
x= (-b(+/-)(b^2-4ac)^1/2)/2a (quadratic equation)
x = (0(+/-)(0^2-((4)(3)(-17))^1/2)/2*3
x = ((+/-)(204)^1/2)/6
x = (+/-)2*((51)^(1/2))/6
x = pos 2/6 * sqrt51 and x = neg 2/6 * sqrt51
for 2y^2 = -5y, rearrange to:
2y^2+5y+0= 0
a=2
b=5
c=0
y = (-5(+/-)(5*5-4*2*0)^(1/2))/(2*2)
y = (-5(+/-)(25)^(1/2))/4
y = (-5 + 5)/4 and (-5 -5)/4
y = 0 and -10/4