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Math  /  Algebra

QuestionThe equation x2+(m−3)x+m=0x^{2}+(m-3) x+m=0 has two real and distinct roots. Determine the range of values for mm.

Studdy Solution

STEP 1

1. The quadratic equation x2+(m−3)x+m=0 x^2 + (m-3)x + m = 0 has real and distinct roots if its discriminant is positive.
2. The discriminant Δ \Delta of a quadratic equation ax2+bx+c=0 ax^2 + bx + c = 0 is given by Δ=b2−4ac \Delta = b^2 - 4ac .
3. For the roots to be real and distinct, Δ>0 \Delta > 0 .

STEP 2

1. Identify the coefficients of the quadratic equation.
2. Write the expression for the discriminant.
3. Set up the inequality for the discriminant to be greater than zero.
4. Solve the inequality to find the range of values for m m .

STEP 3

Identify the coefficients of the quadratic equation x2+(m−3)x+m=0 x^2 + (m-3)x + m = 0 :
- a=1 a = 1 - b=m−3 b = m - 3 - c=m c = m

STEP 4

Write the expression for the discriminant Δ \Delta :
Δ=(m−3)2−4⋅1⋅m \Delta = (m-3)^2 - 4 \cdot 1 \cdot m

STEP 5

Set up the inequality for the discriminant to be greater than zero:
(m−3)2−4m>0 (m-3)^2 - 4m > 0

STEP 6

Expand and simplify the inequality:
(m−3)2−4m=m2−6m+9−4m (m-3)^2 - 4m = m^2 - 6m + 9 - 4m =m2−10m+9>0 = m^2 - 10m + 9 > 0

STEP 7

Solve the quadratic inequality m2−10m+9>0 m^2 - 10m + 9 > 0 .
First, find the roots of the equation m2−10m+9=0 m^2 - 10m + 9 = 0 using the quadratic formula:
m=−b±b2−4ac2a m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
where a=1 a = 1 , b=−10 b = -10 , and c=9 c = 9 .
m=10±(−10)2−4⋅1⋅92⋅1 m = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} m=10±100−362 m = \frac{10 \pm \sqrt{100 - 36}}{2} m=10±642 m = \frac{10 \pm \sqrt{64}}{2} m=10±82 m = \frac{10 \pm 8}{2}
The roots are:
m1=10+82=9 m_1 = \frac{10 + 8}{2} = 9 m2=10−82=1 m_2 = \frac{10 - 8}{2} = 1

STEP 8

Determine the intervals where the inequality m2−10m+9>0 m^2 - 10m + 9 > 0 holds.
The roots divide the number line into intervals: (−∞,1) (-\infty, 1) , (1,9) (1, 9) , and (9,∞) (9, \infty) .
Test each interval to determine where the inequality holds:
- For m<1 m < 1 , choose m=0 m = 0 : 02−10(0)+9=9>0 0^2 - 10(0) + 9 = 9 > 0 - For 1<m<9 1 < m < 9 , choose m=5 m = 5 : 52−10(5)+9=25−50+9=−16<0 5^2 - 10(5) + 9 = 25 - 50 + 9 = -16 < 0 - For m>9 m > 9 , choose m=10 m = 10 : 102−10(10)+9=100−100+9=9>0 10^2 - 10(10) + 9 = 100 - 100 + 9 = 9 > 0
The inequality holds for m∈(−∞,1)∪(9,∞) m \in (-\infty, 1) \cup (9, \infty) .
The range of values for m m is (−∞,1)∪(9,∞) \boxed{(-\infty, 1) \cup (9, \infty)} .

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