a) Two numbers when multiplied equals 36, one number is x and the other number is greater than x by 5 Represent the information given in equation form Express as a quadratic equation Solve the quadratic equation to find possible values for x If told that the two number were positive, give the values of the two original numbers. b) Solve the following quadratic equations ( use the quadratic formula) 1) 2x2−6x−20=0 2) 6x2−5x−6=0

Question

a) Two numbers when multiplied equals 36, one number is x and the other number is greater than x by 5  Represent the information given in equation form  Express as a quadratic equation  Solve the quadratic equation to find possible values for x  If told that the two number were positive, give the values of the two original numbers.  b) Solve the following quadratic equations ( use the quadratic formula)  1) 2x2−6x−20=0  2) 6x2−5x−6=0

giskacu · Accepted Answer

Step 1  Let x be the one numbers.  Other number is greaters than x by 5  So other number is x+5  The two numbers are x and x+5  Product of two numbers is 36  ∴x(x+5)=36  x2+5x=36  x2+5x−36=0  Quadratic equation is x2+5x−36=0  Formula for solving second degree  Quadratic equation is (ax2+bx+c=0)  x=−b±b2−4ac2a  Step 2  x2+5x−36=0  a=1, b=5, c=−36  x=−5±52−4x1x−36{2x1}−−5±25+1442  x=−5±1692=−5±1692  x=−5+132 or x=−5−132  x=4 or x=−9  Possible values of x are x=4 and x=−9  Given that two numbers were positive  So the value of x is 4  Other number is x+5=4+5=9  So the numbers are 4 and 9  Step 3  1) 2x2−6x−20=0  x=−b±b2−4ac2a=−(−6)±(−6)2−4x×2×−202×2  =6±36+1604=6±144  x=6+144=5 or x=6−144=−2  Possible values of x are 5 and -2  2) 6x2−5x−6=0  x=−(−5)±(−5)2−4×6×−62×6  =5±25+14412=5±1312  x=5+

a) Two numbers when multiplied equals 36, one number is x and the othe

Answered question

Answer & Explanation

New Questions in Algebra I