PLEASE EXPLAIN HOW TO DO THIS PROBLEM STEP BY STEP! THANK SOOMUCH!!! 1.) Which o

Question

PLEASE EXPLAIN HOW TO DO THIS PROBLEM STEP BY STEP! THANK SOOMUCH!!! 1.) Which of the following is false for some real numberx? a.) x^2-x+1>0 b.) x=x^2 c.) 1=x^2+1/x^2+1 d.) x=cubed root ofx^3 PLEASE EXPLAIN HOW TO DO THIS PROBLEM STEP BY STEP! THANK SOOMUCH!!! 1.) Which of the following is false for some real numberx? a.) x^2-x+1>0 b.) x=x^2 c.) 1=x^2+1/x^2+1 d.) x=cubed root ofx^3

Explanation / Answer

Well, first you have to understand what "false for real number x"means. It simply means for what values of x will the functionsbelow not be true. Example x+3=5 then x=7 will be false. Only x=2will make it true. Similarly: a.) x^2-x+1>0 In other words for what value of x will the expression on the leftbe less than zero or equal to zero! (i.e. it will defy thecondition >0). Well if we try equating to zero and then try to factorize using thequadratic equation we will see that there are no real roots!Therefore we can't make the left hand side = 0. On the other hand, can we make it less than zero? Well from x^2-x+1we can see that it is impossible to obtain zero as whatever valuewe use as x, x^2 will always be a bigger positive that than the xwe minus thereafter. Thus, a) is not false for some real number ofx. b.) x=x^2 This one is easy. If x is positive number it works fine, since theroot of the square of a number will always be equal to that numberbut only if the number is positive. However if x is negative, thenit won't work. Example: x=-5 -5=(-5)^2 = 25=5 and this isobviously not correct as -5 does not equal 5! Therefore b) is false for all negative real numbers of x. c.) 1=x^2+1/x^2+1 Here we must check if the denominator can be zero.x^2+1=0   ==> x^2=-1 and we cannot find the squareroot of -1 on the real number line therefore the denominator canalways be zero. There is no other way that the expression can beproven false using a real number as x so c) is not false for somereal number of x. d.) x=cubed root ofx^3 This is similar to b) but the cube root caters for negative numbersi.e. x=-3 so -3=cube (-3)^3 = cube -27= -3. Therefored) is not false for some real number of x.

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