Finding the minimum value.
$begingroup$
I'm struck on this question, I tried hard but couldn't solve it.
Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.
So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.
Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$
Any help will be appreciated.
calculus algebra-precalculus quadratics quadratic-forms
edited Dec 8 '18 at 12:22
amWhy
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
$endgroup$
add a comment |
$begingroup$
I'm struck on this question, I tried hard but couldn't solve it.
Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.
So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.
Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$
Any help will be appreciated.
calculus algebra-precalculus quadratics quadratic-forms
edited Dec 8 '18 at 12:22
amWhy
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
$endgroup$
add a comment |
$begingroup$
I'm struck on this question, I tried hard but couldn't solve it.
Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.
So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.
Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$
Any help will be appreciated.
calculus algebra-precalculus quadratics quadratic-forms
edited Dec 8 '18 at 12:22
amWhy
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
$endgroup$
I'm struck on this question, I tried hard but couldn't solve it.
Question: if a quadratic equation in $x$: $$ax^2 - bx + 5 = 0$$ does not have two distinct real roots, then find the minimum value of $5a + b$.
So far, I tried using the condition that the discriminant should be negative or zero, but couldn't proceed further.
Moreover, as the given equation doesn't have two distinct roots so the graph will be either concave upwards or downwards, by double differentiation, I found that graph will be concave upwards so this equation will be positive or zero for all real $x.$
Any help will be appreciated.
calculus algebra-precalculus quadratics quadratic-forms
calculus algebra-precalculus quadratics quadratic-forms
edited Dec 8 '18 at 12:22
amWhy
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
edited Dec 8 '18 at 12:22
amWhy
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
edited Dec 8 '18 at 12:22
amWhy
1
edited Dec 8 '18 at 12:22
amWhy
1
edited Dec 8 '18 at 12:22
amWhy
1
1
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
asked Dec 8 '18 at 12:15
Shivansh JShivansh J
346
346
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$Dleq0$$
$$Longrightarrow b^2leq20a$$
$$Longrightarrow 5ageqfrac{b^2}{4}$$
$$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
where equality holds if $b = -2 and a = frac{1}{5}$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.
Hope it is helpful:)
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
$endgroup$
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$$Dleq0$$
$$Longrightarrow b^2leq20a$$
$$Longrightarrow 5ageqfrac{b^2}{4}$$
$$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
where equality holds if $b = -2 and a = frac{1}{5}$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.
Hope it is helpful:)
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
$endgroup$
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
add a comment |
$begingroup$
$$Dleq0$$
$$Longrightarrow b^2leq20a$$
$$Longrightarrow 5ageqfrac{b^2}{4}$$
$$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
where equality holds if $b = -2 and a = frac{1}{5}$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.
Hope it is helpful:)
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
$endgroup$
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
add a comment |
$begingroup$
$$Dleq0$$
$$Longrightarrow b^2leq20a$$
$$Longrightarrow 5ageqfrac{b^2}{4}$$
$$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
where equality holds if $b = -2 and a = frac{1}{5}$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.
Hope it is helpful:)
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
$endgroup$
$$Dleq0$$
$$Longrightarrow b^2leq20a$$
$$Longrightarrow 5ageqfrac{b^2}{4}$$
$$Longrightarrow 5a+bgeq frac{b^2}{4}+bgeq -1$$
where equality holds if $b = -2 and a = frac{1}{5}$
Also, you need not double differentiate to get that the graph is concave upwards, just see that it is taking positive value at 0, so it will take non-negative values for all real $x$, hence, it must be concave upwards.
Hope it is helpful:)
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
edited Dec 19 '18 at 13:50
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
answered Dec 8 '18 at 12:49
MartundMartund
1,593212
1,593212
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
add a comment |
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
1
1
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
$begingroup$
Good answer! Just to make things clearer, that last $frac{b^2}{4}+bgeq -1$ can be easily checked by completing the square.
$endgroup$
– s0ulr3aper07
Dec 8 '18 at 12:59
1
1
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Yes, that is how I did it.
$endgroup$
– Martund
Dec 8 '18 at 13:01
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
$begingroup$
Thanks for help..finally I know how to deal with these problems.
$endgroup$
– Shivansh J
Dec 8 '18 at 16:03
add a comment |
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