Root 2x square +7x+5 root 2 .solve for x
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root2x^2+7×+5root2
root2x^2+5×+2×+5root2
root2x (root2x+5×)+root2 (root2x+5)
(root2x+root2)(root2x+5)
root2x=-5;root2x=-root2
x=-5/root2;x=-root2/root 2
it is not ;x=-1
possible for
this
.°.X=-1
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