To prove the relation between the roots of three quadratic equationsIf x1 is a root of ax2+bx+c=0,x2 is a root of −ax2+bx+c=0 where 0&lt;x1&lt;x2, show that the equation ax2+2bx+2c=0 has a root satisfying 0&lt;x1&lt;x3&lt;x2.

Given:ax12+bx1+c==0≈∼(1),≈−ax22+bx2+c=0,≈(2)Therefore, f(x)=ax2+2bx+2c⇒f(x1)=ax12+2bx1+2c≈∼(3),≈∼f(x2)=ax22+2bx2+2c≈∼(4)Use (1) in (4) to get f(x1)=−ax12 and (2) in (4) to get f(x2)=3ax22 Sof(x1)f(x2)=−3a2x12x22&lt;0Therefore by IVT there exists x3∈(x1,x2) such that f(x3)=0.

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High School
Algebra
Algebra I
Quadratic function and equation

daurialebraslmc

Answered question

2022-04-11

To prove the relation between the roots of three quadratic equations
If

x_{1}

is a root of

a x^{2} + b x + c = 0, x_{2}

is a root of

- a x^{2} + b x + c = 0

where

0 < x_{1} < x_{2}

, show that the equation

a x^{2} + 2 b x + 2 c = 0

has a root satisfying

0 < x_{1} < x_{3} < x_{2}

Answer & Explanation

Kendall Wilkinson

Beginner2022-04-12Added 17 answers

Given:

a x_{1}^{2} + b x_{1} + c == 0 \approx\sim (1), \approx - a x_{2}^{2} + b x_{2} + c = 0, \approx (2)

Therefore,

f (x) = a x^{2} + 2 b x + 2 c \Rightarrow f (x_{1}) = a x_{1}^{2} + 2 b x_{1} + 2 c \approx\sim (3), \approx\sim f (x_{2}) = a x_{2}^{2} + 2 b x_{2} + 2 c \approx\sim (4)

Use (1) in (4) to get

f (x_{1}) = - a x_{1}^{2}

and (2) in (4) to get

f (x_{2}) = 3 a x_{2}^{2}

f (x_{1}) f (x_{2}) = - 3 a^{2} x_{1}^{2} x_{2}^{2} < 0

Therefore by IVT there exists

x_{3} \in (x_{1}, x_{2})

such that

f (x_{3}) = 0 .

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