0000:00:00:00:00Start2Number 2ExplanationIf x∗y=(x+y+1)x+xyx \ast y = (x + y + 1)x + xyx∗y=(x+y+1)x+xy and a∗(−1∗2)=−10a \ast (-1 \ast 2) = -10a∗(−1∗2)=−10, then the value of aaa satisfying the equation is ....222 or −5-5−5222 or 555444 or −2-2−2−2-2−2 or 555−4-4−4 or −2-2−2ExplanationGiven the operation x∗y=(x+y+1)x+xyx \ast y = (x + y + 1)x + xyx∗y=(x+y+1)x+xy. First, we calculate the value of −1∗2-1 \ast 2−1∗2. (−1∗2)=(−1+2+1)(−1)+(−1)(2)(-1 \ast 2) = (-1 + 2 + 1)(-1) + (-1)(2)(−1∗2)=(−1+2+1)(−1)+(−1)(2)=(2)(−1)−2= (2)(-1) - 2=(2)(−1)−2=−2−2= -2 - 2=−2−2=−4= -4=−4 Next, substitute this result into the equation a∗(−1∗2)=−10a \ast (-1 \ast 2) = -10a∗(−1∗2)=−10, which becomes a∗(−4)=−10a \ast (-4) = -10a∗(−4)=−10. a∗(−4)=(a+(−4)+1)a+a(−4)=−10a \ast (-4) = (a + (-4) + 1)a + a(-4) = -10a∗(−4)=(a+(−4)+1)a+a(−4)=−10(a−3)a−4a=−10(a - 3)a - 4a = -10(a−3)a−4a=−10a2−3a−4a=−10a^2 - 3a - 4a = -10a2−3a−4a=−10a2−7a+10=0a^2 - 7a + 10 = 0a2−7a+10=0 We factor the quadratic equation to find the value of aaa. (a−2)(a−5)=0(a - 2)(a - 5) = 0(a−2)(a−5)=0a=2ora=5a = 2 \quad \text{or} \quad a = 5a=2ora=5 Thus, the values of aaa that satisfy the equation are 222 or 555.
2Number 2ExplanationIf x∗y=(x+y+1)x+xyx \ast y = (x + y + 1)x + xyx∗y=(x+y+1)x+xy and a∗(−1∗2)=−10a \ast (-1 \ast 2) = -10a∗(−1∗2)=−10, then the value of aaa satisfying the equation is ....222 or −5-5−5222 or 555444 or −2-2−2−2-2−2 or 555−4-4−4 or −2-2−2ExplanationGiven the operation x∗y=(x+y+1)x+xyx \ast y = (x + y + 1)x + xyx∗y=(x+y+1)x+xy. First, we calculate the value of −1∗2-1 \ast 2−1∗2. (−1∗2)=(−1+2+1)(−1)+(−1)(2)(-1 \ast 2) = (-1 + 2 + 1)(-1) + (-1)(2)(−1∗2)=(−1+2+1)(−1)+(−1)(2)=(2)(−1)−2= (2)(-1) - 2=(2)(−1)−2=−2−2= -2 - 2=−2−2=−4= -4=−4 Next, substitute this result into the equation a∗(−1∗2)=−10a \ast (-1 \ast 2) = -10a∗(−1∗2)=−10, which becomes a∗(−4)=−10a \ast (-4) = -10a∗(−4)=−10. a∗(−4)=(a+(−4)+1)a+a(−4)=−10a \ast (-4) = (a + (-4) + 1)a + a(-4) = -10a∗(−4)=(a+(−4)+1)a+a(−4)=−10(a−3)a−4a=−10(a - 3)a - 4a = -10(a−3)a−4a=−10a2−3a−4a=−10a^2 - 3a - 4a = -10a2−3a−4a=−10a2−7a+10=0a^2 - 7a + 10 = 0a2−7a+10=0 We factor the quadratic equation to find the value of aaa. (a−2)(a−5)=0(a - 2)(a - 5) = 0(a−2)(a−5)=0a=2ora=5a = 2 \quad \text{or} \quad a = 5a=2ora=5 Thus, the values of aaa that satisfy the equation are 222 or 555.
