X^2 + 11x= -28 Solutions of quadratic equation
\[x^2+11x+28=0\] 1*28=28 4+7=11 4*7=28 write 11x=4x+7x and make factors.
To pan it out a bit more. You need to get all the numbers on 1 side so the equation is equal to 0. So you add 28 to boths sides to get \[x ^{2} +11x + 28 = 0\] Now we need to find a pair of numbers that add to make 28 and multiply to make 11. As surji pointed out, 7x4 = 28 and they also add to make 11 so 7 and 4 and the numbers we want. Now we can because the X^2 term has a coefficent we can put these numbers straight into factors. (x+7)(x+4). This bracked when multiplied out will give the quadratic sown above. But beacuse the equation is = to 0 we now need to work out what x has to be for this to equal 0. So (x+7)(x+4) = 0 If x was -7 let's say, then we would end up with 0 x -3 and 0 times anything = 0 so -7 works. and if x was -4 the same would happen so -4 and -7 are the answers.
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