Question

Find all values of m for which the equation has two real solutions.

3x² + 7x - (m + 1) = 0
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Answer 1

Given equation to us is ,
`3x^2+7x-(m+1)=0` .

The given equation is a quadratic equation and it has two real solutions if it's discriminant is greater than or equal to 0 .

For a quadratic equation in standard form of
`ax ^(2) + bx + c = 0`

the discriminant is given by b² - 4ac .

On comparing wrt to Standard form, we have ;

• 
  `a = 3`
• 
  `b = 7`
• 
  `c = - (m+1)`

Hence we have,

`\longrightarrow b^2-4ac \geq 0\\`

`\longrightarrow 7^2-4(3)-(m+1)\geq 0\\`

`\longrightarrow 49 +12(m+1)\geq 0\\`

`\longrightarrow 49 +12m+12\geq 0 \\`

`\longrightarrow 12m \geq -61\\`

`\longrightarrow \underline{\underline{m \geq (-61)/(12)}}`

Therefore the given quadratic equation will have real solutions for values of m greater than or equal to -61/12 .

And we are done!