Applying quadratic equations.
$begingroup$
So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.
This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.
So my solution is:
side = $x$
length = $2x$
so then:
length x width = length + length + width + width + $3.5 $
$2x*x = 2x + 2x + x + x + 3.5 $
$2x^2 = 6x + 3.5 $
$-2x^2 + 6x + 3.5 = 0$
Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.
quadratics
asked Jul 14 '18 at 6:10
MEchoMEcho
11
$endgroup$
|
show 2 more comments
$begingroup$
So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.
This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.
So my solution is:
side = $x$
length = $2x$
so then:
length x width = length + length + width + width + $3.5 $
$2x*x = 2x + 2x + x + x + 3.5 $
$2x^2 = 6x + 3.5 $
$-2x^2 + 6x + 3.5 = 0$
Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.
quadratics
asked Jul 14 '18 at 6:10
MEchoMEcho
11
$endgroup$
4
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06
|
show 2 more comments
$begingroup$
So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.
This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.
So my solution is:
side = $x$
length = $2x$
so then:
length x width = length + length + width + width + $3.5 $
$2x*x = 2x + 2x + x + x + 3.5 $
$2x^2 = 6x + 3.5 $
$-2x^2 + 6x + 3.5 = 0$
Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.
quadratics
asked Jul 14 '18 at 6:10
MEchoMEcho
11
$endgroup$
So I realise this is quite an easy question, but for some reason I can't see the solution. So the question followed on from a previous question where we used a quadratic equation to find the dimensions of a right angle triangle.
This question is: Find the dimensions of all rectangles in which the area equals the perimeter + $3.5$, and in which the longer sides are twice the length of the shorter sides.
So my solution is:
side = $x$
length = $2x$
so then:
length x width = length + length + width + width + $3.5 $
$2x*x = 2x + 2x + x + x + 3.5 $
$2x^2 = 6x + 3.5 $
$-2x^2 + 6x + 3.5 = 0$
Then I would factorise this to find the positive values of $x$ that can be then used to determine the length of the rectangle. I am struggling to factorise while one of the values is $3.5$. Is there a way to more easily conceptualise how to factorise with non whole values? thanks.
quadratics
quadratics
asked Jul 14 '18 at 6:10
MEchoMEcho
11
asked Jul 14 '18 at 6:10
MEchoMEcho
11
edited Jul 14 '18 at 6:17
user570193
asked Jul 14 '18 at 6:10
MEchoMEcho
11
asked Jul 14 '18 at 6:10
MEchoMEcho
11
asked Jul 14 '18 at 6:10
MEchoMEcho
11
11
4
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06
|
show 2 more comments
4
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06
4
4
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "
Yeah, it's called completing the square or the quadratic formula.
If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....
$ax^2 + bx + c = 0$
$x^2 + frac ba = -frac ca$
$x^2 + 2frac b{2a} = - frac ca$
$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$
$(x + frac b{2a})^2 + frac {b^2}{4a^2} -frac {4ac}{4a^2}=frac {b^2 - 4ac}{4a^2}$
$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$
$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.
So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $
$= frac {-6 pm sqrt{36 + 28}}{-4}$
$= frac {6 mp sqrt{64}}{4} $
$= frac {6 mp 8}{4} $
$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so
the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.
And .... indeed, $24frac 12 = 21 + 3.5$.
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
$$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
$endgroup$
add a comment |
$begingroup$
$$2x^2 - 6x - 3.5 = 0$$
I wondered if the $ac$ method would still work. It sort of can.
$ac =2(-3.5) = -7$ and $b=-6$.
We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.
So we replace $bx=-6x$ with $-6x = -7x + 1x$.
begin{align}
2x^2 - 6x - 3.5
&= 2x^2 -7x + 1x - 3.5 \
&= 2x(x-3.5) + 1(x-3.5) \
&= (2x+1)(x-3.5) \
hline
x &in {-0.5, 3.5}
end{align}
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "
Yeah, it's called completing the square or the quadratic formula.
If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....
$ax^2 + bx + c = 0$
$x^2 + frac ba = -frac ca$
$x^2 + 2frac b{2a} = - frac ca$
$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$
$(x + frac b{2a})^2 + frac {b^2}{4a^2} -frac {4ac}{4a^2}=frac {b^2 - 4ac}{4a^2}$
$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$
$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.
So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $
$= frac {-6 pm sqrt{36 + 28}}{-4}$
$= frac {6 mp sqrt{64}}{4} $
$= frac {6 mp 8}{4} $
$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so
the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.
And .... indeed, $24frac 12 = 21 + 3.5$.
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "
Yeah, it's called completing the square or the quadratic formula.
If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....
$ax^2 + bx + c = 0$
$x^2 + frac ba = -frac ca$
$x^2 + 2frac b{2a} = - frac ca$
$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$
$(x + frac b{2a})^2 + frac {b^2}{4a^2} -frac {4ac}{4a^2}=frac {b^2 - 4ac}{4a^2}$
$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$
$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.
So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $
$= frac {-6 pm sqrt{36 + 28}}{-4}$
$= frac {6 mp sqrt{64}}{4} $
$= frac {6 mp 8}{4} $
$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so
the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.
And .... indeed, $24frac 12 = 21 + 3.5$.
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
$endgroup$
add a comment |
$begingroup$
"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "
Yeah, it's called completing the square or the quadratic formula.
If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....
$ax^2 + bx + c = 0$
$x^2 + frac ba = -frac ca$
$x^2 + 2frac b{2a} = - frac ca$
$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$
$(x + frac b{2a})^2 + frac {b^2}{4a^2} -frac {4ac}{4a^2}=frac {b^2 - 4ac}{4a^2}$
$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$
$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.
So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $
$= frac {-6 pm sqrt{36 + 28}}{-4}$
$= frac {6 mp sqrt{64}}{4} $
$= frac {6 mp 8}{4} $
$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so
the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.
And .... indeed, $24frac 12 = 21 + 3.5$.
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
$endgroup$
"Is there a way to more easily conceptualise how to factorise with non whole values? thanks. "
Yeah, it's called completing the square or the quadratic formula.
If you have $ax^2 + bx + c = 0$ and $a ne 0$ then.....
$ax^2 + bx + c = 0$
$x^2 + frac ba = -frac ca$
$x^2 + 2frac b{2a} = - frac ca$
$x^2 + 2frac b{2a} + (frac b{2a})^2 = (frac b{2a})^2 - frac ca$
$(x + frac b{2a})^2 + frac {b^2}{4a^2} -frac {4ac}{4a^2}=frac {b^2 - 4ac}{4a^2}$
$x + frac b{2a} = pm sqrt {frac {b^2 - 4ac}{4a^2}} = frac {sqrt{b^2 - 4ac}}{2a}$
$x = frac {-b pm sqrt {b^2 - 4ac}}{2a}$.
So in your case $x = frac {-6 pm sqrt{6^2 - 4*3.5*(-2)}}{2*(-2)} $
$= frac {-6 pm sqrt{36 + 28}}{-4}$
$= frac {6 mp sqrt{64}}{4} $
$= frac {6 mp 8}{4} $
$-frac 12$ or $frac 72$. As $x > 0$ we have $x = frac 72$ so
the sides are $frac 72$ and $7$ and area is $frac 72*7 = frac {49}2 = 24frac 12$ and the perimeter is $frac 72 + frac 72 + 7 + 7 = 21$.
And .... indeed, $24frac 12 = 21 + 3.5$.
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
answered Jul 14 '18 at 7:27
fleabloodfleablood
72k22687
72k22687
add a comment |
add a comment |
$begingroup$
Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
$$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
$endgroup$
add a comment |
$begingroup$
Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
$$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
$endgroup$
add a comment |
$begingroup$
Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
$$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
$endgroup$
Without applying (or re-inventing) the formulæ, multiply your equation by $2 $ to rewrite it as
$$4x^2-12x-7=(2x-3)^2-9-7=0iff (2x-3)^2=4^2$$
so the longer size is $2x=4+3=7:$ and the shorter size is $:x=7/2$.
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
answered Jul 14 '18 at 8:01
BernardBernard
122k741116
122k741116
add a comment |
add a comment |
$begingroup$
$$2x^2 - 6x - 3.5 = 0$$
I wondered if the $ac$ method would still work. It sort of can.
$ac =2(-3.5) = -7$ and $b=-6$.
We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.
So we replace $bx=-6x$ with $-6x = -7x + 1x$.
begin{align}
2x^2 - 6x - 3.5
&= 2x^2 -7x + 1x - 3.5 \
&= 2x(x-3.5) + 1(x-3.5) \
&= (2x+1)(x-3.5) \
hline
x &in {-0.5, 3.5}
end{align}
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
$endgroup$
add a comment |
$begingroup$
$$2x^2 - 6x - 3.5 = 0$$
I wondered if the $ac$ method would still work. It sort of can.
$ac =2(-3.5) = -7$ and $b=-6$.
We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.
So we replace $bx=-6x$ with $-6x = -7x + 1x$.
begin{align}
2x^2 - 6x - 3.5
&= 2x^2 -7x + 1x - 3.5 \
&= 2x(x-3.5) + 1(x-3.5) \
&= (2x+1)(x-3.5) \
hline
x &in {-0.5, 3.5}
end{align}
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
$endgroup$
add a comment |
$begingroup$
$$2x^2 - 6x - 3.5 = 0$$
I wondered if the $ac$ method would still work. It sort of can.
$ac =2(-3.5) = -7$ and $b=-6$.
We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.
So we replace $bx=-6x$ with $-6x = -7x + 1x$.
begin{align}
2x^2 - 6x - 3.5
&= 2x^2 -7x + 1x - 3.5 \
&= 2x(x-3.5) + 1(x-3.5) \
&= (2x+1)(x-3.5) \
hline
x &in {-0.5, 3.5}
end{align}
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
$endgroup$
$$2x^2 - 6x - 3.5 = 0$$
I wondered if the $ac$ method would still work. It sort of can.
$ac =2(-3.5) = -7$ and $b=-6$.
We note that $(-7)(1) = -7 = ac$ and $(-7)+(1) = -6 = b$.
So we replace $bx=-6x$ with $-6x = -7x + 1x$.
begin{align}
2x^2 - 6x - 3.5
&= 2x^2 -7x + 1x - 3.5 \
&= 2x(x-3.5) + 1(x-3.5) \
&= (2x+1)(x-3.5) \
hline
x &in {-0.5, 3.5}
end{align}
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
edited Jan 3 at 13:55
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
answered Jan 3 at 7:37
steven gregorysteven gregory
18.3k32258
18.3k32258
add a comment |
add a comment |
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4
$begingroup$
Multiply by $2$
$endgroup$
– Claude Leibovici
Jul 14 '18 at 6:14
$begingroup$
Do you mean multiply every term by two,or? Because multiplying 3.5 gives 7, not six.
$endgroup$
– MEcho
Jul 14 '18 at 6:16
$begingroup$
Note that $2cdot 7-2cdot 1=12$... (Yes, you would multiply through the whole quadratic by $2$.)
$endgroup$
– abiessu
Jul 14 '18 at 6:20
$begingroup$
Oh okay, I see now! thanks!
$endgroup$
– MEcho
Jul 14 '18 at 6:21
$begingroup$
Use the quadratic formula.
$endgroup$
– MalayTheDynamo
Jul 14 '18 at 7:06