How do you solve the following with the quadratic formula?: #sqrt2x² - x - 3sqrt(2) =0#

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Answer 1 · 2018-07-08T15:31:29

# x=(3sqrt2)/2, or, x=-sqrt2#.

Comparing the given quadratic equation with the standard one, i.e.,

#ax^2+bx+c=0#, we have,

#a=sqrt2, b=-1, and c=-3sqrt2#.

As per the quadratic formula, the roots are,

#x={-a+-sqrtDelta}/(2a)," where, "Delta=(b^2-4ac)#.

In our case, #Delta=(-1)^2-4(sqrt2)(-3sqrt2)#,

#=1+24#.

# :. Delta=25," so that, "sqrtDelta=5#.

Hence, #-b+-sqrtDelta=1+-5=6, or, -4#.

Finally, we get the roots : #6/(2sqrt2), or, -4/(2sqrt2),#

# i.e., 3/sqrt2=(3sqrt2)/2, or, -sqrt2#.

Answer 2 · 2018-07-08T15:33:46

#x=-sqrt2,(3sqrt2)/2#

Given: #sqrt2x^2-x-3sqrt2=0#.

Use the quadratic formula, which states that,

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Here, #a=sqrt2,b=-1,c=-3sqrt2#.

#:.x=(1+-sqrt(1-4*sqrt2*-3sqrt2))/(2sqrt2)#

#=(1+-sqrt(1+24))/(2sqrt2)#

#=(1+-sqrt25)/(2sqrt2)#

#=(1+-5)/(2sqrt2)#

#:.x_1=(1+5)/(2sqrt2)=6/(2sqrt2)=3/sqrt2=(3sqrt2)/2#

#:.x_2=(1-5)/(2sqrt2)=-4/(2sqrt2)=-2/sqrt2=-sqrt2#