Best solutions to - "Find m such that x4−(2m−1)x2+4m−5=0 has real roots"

High School
Algebra
Algebra I
Quadratic function and equation

Elijah Schwartz

Answered question

2022-04-12

Find m such that

x^{4} - (2 m - 1) x^{2} + 4 m - 5 = 0

has real roots

Answer & Explanation

embemiaEffoset4rs

Beginner2022-04-13Added 14 answers

Step 1
Put

m = 2

. Then u is not real, so x is not real. Instead, if

m \geq \frac{5}{4}

write out the two quadratic factors and use the condition for them(it's the same condition for each factor) to have real roots.After simpplification, you finally get

(2 m - 7) (2 m - 3) \geq 0

1.25 \leq m \leq 1.5

m \geq 3.5

superpms01wks1

Beginner2022-04-14Added 13 answers

Step 1
You need also to consider

Δ

to be positive in order for the solutions to be real.

Δ = {(2 m - 1)}^{2} - 4 (4 m - 5) = 4 m^{2} - 4 m + 1 - 16 m + 20 = 4 m^{2} - 20 m + 21

m_{1, 2} = \frac{10 \pm \sqrt{100 - 84}}{4} = \frac{10 \pm 4}{4} = {\frac{3}{2}, \frac{7}{2}}

Thus

Δ \geq 0 \Leftrightarrow m \in (- \infty, \frac{3}{2}] \cup [\frac{7}{2}, \infty)

Take for example

m = 2

: now

m \geq \frac{5}{4}

but the equation

u^{2} - (2 \cdot 2 - 1) u + 4 \cdot 2 - 5 = 0

u^{2} - 3 u + 3 = 0

has not real solutions

\frac{3 \pm \sqrt{- 3}}{2}

Hence we need

m \in [\frac{7}{2}, \infty)

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