Question

For all x, x²+2ax +10-3a> 0, then the interval in which a lies is

(A)a <-5

(B)-5 < a < 2

(C)a> 5

(D) 2<a<5



say the correct answer if the answer is correct I'll mark u brainliest​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

For all x, x² +2ax + 10 - 3a > 0, then the interval in which a lies is

(A)a <-5

(B)-5 < a < 2

(C)a> 5

(D) 2<a<5

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$\huge \orange{AηsωeR} ✍$

$\large\underline\blue{\bold{Given :-  }}$

For all x, x² +2ax + 10 - 3a > 0

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$\large\underline\purple{\bold{To Find :- }}$

The interval in which a lies is

(A)a <-5

(B)-5 < a < 2

(C)a> 5

(D) 2<a<5

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$\bf \:\large\underline\purple{\bold{Understanding \: the \: Concept :- }}$

Let us consider a quadratic polynomial f(x) = ax² +bx + c , then f(x) > 0, is possible only when

$\sf \:  ⟼(1). \: a > 0$

$\sf \:  ⟼(2). \: {b}^{2} - 4ac > 0$

$\sf \:  ⟼(3).(x - a)(x - b) > 0\bf\implies \:x < a \: or \: x > b$

$\sf \:  provided \: that \: a < b$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \:For all x, \: x² +2ax + 10 - 3a > 0,$

$\bf\implies \: {(2a)}^{2} - 4 \times 1 \times (10 - 3a) > 0$

$\bf\implies \: {4a}^{2} - 40 + 12a > 0$

$\bf\implies \: {4a}^{2} + 12a - 40 > 0$

$\bf\implies \: {4a}^{2} + 20a - 8a - 40 > 0$

$\bf\implies \: 4a(a + 5) - 8(a + 5) > 0$

$\bf\implies \:(a + 5)(4a - 8) > 0$

$\bf\implies \:4(a + 5)(a - 2) > 0$

$\bf\implies \:a < - 5 \: or \: a > 2$

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$\large{\boxed{\boxed{\bf{Option (a) \: is \: correct}}}}$

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Answer 2

ᴀɴsᴡᴇʀ

∴−5<a<2

Given equation-

sᴛᴇᴘ ʙʏ sᴛᴇᴘ sᴘᴏʟᴜᴛɪᴏɴ

x2+2ax+(10−3a)>0 for all x∈R.

Here, A=1,B=2a,C=(10−3a)

As we know that Ax2+Bx+C>0 for all x∈R if-

A>0 and D<0

A=1>0

Now,

D<0

B2−4AC<0

(2a)2−4(1)(10−3a)<0

4a2−40+12a<0

a2+3a−10=0

(a+5)(a−2)<0

a∈(−5,2)

∴−5<a<2