If (m+8)(x^2)+m=7-8x has two real roots show that (m+9)(m-8)<0 where m is an arbitrary constant
For this we are going to test our knowledge of discriminats and factorisation. Firstly we will format the equation as (m+8)x2+8x+(m-7)=0From here we can see it takes the form ax2+bx+c, and as we know the equation has two real roots we know that the discriminant D is greater than 0. Therfore b2-4ac>0a=(m+8) b=8 c=(m-7)82-4(m+8)(m-7)>0 Pluggin in a,b,c64-4(m2+m-54)>0 expanding brackets and squaring 816-(m2+m-54)>0 dividing both sides by a factor of 40>(m2+m-54)-16 moving left handside to right handside by addition / subtraction0>m2+m-72 collecting terms0>(m+9)(m-8) factorisingWe have now shown that (m+9)(m-8)<0 for the above equation when it has two real roots.
MJ
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