Question

If the length of HJ is 7 find x

Answer 1

So to find x we need to add both HI and IJ - which we know equals 7:

$(4x+13)+(x^{2}+3x+4)=7$

Then we need to combine like terms:

$x^{2}+7x+17=7$

And then set equal to 0:

$x^{2}+7x+10=0$

Then we can factor:

$(x+5)(x+2)=0$

So then when we set each term equal to 0 individually and solve for x we get: x = -5 and x = -2.

So we have these two values - we need to test which one is correct. So if we go back to the HI segment, let's try plugging in x = -5:

$4(-5)+13=-7$

This gives us a negative number, which we know that length cannot be negative. Let's try x = -2:

$4(-2)+13=5$

This gives us a positive length and therefore we can keep this value.

So the answer is: x = -2.

Answer 2

HJ = HI + IJ

HJ = (4x+13) +(x² + 3x + 4)= x² + 7x + 17, also HJ = 7

x² + 7x + 17=7

x² + 7x +10=0

(x+5)(x+2)=0

x=-5, or x= - 2

Check:

HJ = x² + 7x + 17

x= - 5, HI = 4x + 13= -4*5 + 13 = -7 (It is impossible, because length of the segment cannot be negative).

x= -2, HI= 4x + 13= 4*(-2) +13 = 5

Answer: x is -2.