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

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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)=0

From 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>0

a=(m+8) b=8 c=(m-7)

8²-4(m+8)(m-7)>0 Pluggin in a,b,c

64-4(m²+m-54)>0 expanding brackets and squaring 8

16-(m²+m-54)>0 dividing both sides by a factor of 4

0>(m²+m-54)-16 moving left handside to right handside by addition / subtraction

0>m²+m-72 collecting terms

0>(m+9)(m-8) factorising

We have now shown that (m+9)(m-8)<0 for the above equation when it has two real roots.

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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

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