Joana solved the equation 4x2−4x−3=0%0D%0A4%0D%0A%0D%0A2%0D%0A−%0D%0A4%0D%0A%0D%0A−%0D%0A3%0D%0A=%0D%0A0%0D%0A . She followed the steps to solve a quadratic equation by grouping. However, in Step 3 she noticed that the two groupings did not share a common binomial, as shown in her work below. She concluded that the quadratic equation 4x2−4x−3=0%0D%0A4%0D%0A%0D%0A2%0D%0A−%0D%0A4%0D%0A%0D%0A−%0D%0A3%0D%0A=%0D%0A0%0D%0A has no solutions. Is Joana’s conclusion correct? Review Joana’s work to answer the question. %0D%0A%0D%0AStep 1: 4x2+2x+(−6x)−3=0%0D%0A4%0D%0A%0D%0A2%0D%0A+%0D%0A2%0D%0A%0D%0A+%0D%0A(%0D%0A−%0D%0A6%0D%0A%0D%0A)%0D%0A−%0D%0A3%0D%0A=%0D%0A0%0D%0A %0D%0A%0D%0AStep 2: (4x2+2x)+(−6x−3)=0%0D%0A(%0D%0A4%0D%0A%0D%0A2%0D%0A+%0D%0A2%0D%0A%0D%0A)%0D%0A+%0D%0A(%0D%0A−%0D%0A6%0D%0A%0D%0A−%0D%0A3%0D%0A)%0D%0A=%0D%0A0%0D%0A %0D%0A%0D%0AStep 3: x(4x+2)+(−3(2x+1))=0
Reviewing Joana's work:
Step 1: 4x2+2x+(−6x)−3=0
Step 2: (4x2+2x)+(−6x−3)=0
Step 3: x(4x+2)+(−3(2x+1))=0
In Step 3, Joana applied the distributive property correctly to simplify the expression. However, there is an error in her statement that the two groupings do not share a common binomial.
The expression x(4x+2) and (-3(2x+1)) both have a common binomial of (2x+1). Thus, Joana's conclusion that the quadratic equation has no solutions is incorrect.
To continue solving the equation, Joana could apply the zero product property to set each grouping equal to zero and solve for x:
x(4x+2) = 0 -> x = 0 or 4x + 2 = 0
-3(2x+1) = 0 -> -6x-3 = 0 -> -6x = 3 -> x = -1/2