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Let 7x^2 + 5x = h + 3x^2 - 11x.  What value of will h give us exactly one solution for x?

Aug 15, 2021

#1
+31
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Aug 15, 2021
#2
+780
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4x^2+16x-h=0

4(x^2 + 4x - x/4) = 0

x^2 + 15x/4=0

x = - 15/4

Aug 15, 2021
#3
+12222
+1

Let 7x^2 + 5x = h + 3x^2 - 11x.  What value of will h give us exactly one solution for x?

Hello Guest!

$$4x^2+16x-h=0\\ x^2+4x-\frac{h}{4}=0\\ x=-2\pm\sqrt{4+\frac{h}{4}}$$

$$4+\dfrac{h}{4}=0\\ 4=-\dfrac{h}{4}\\ \color{blue}h=-16$$

h = - 16 will  give us exactly one solution for x.

!

Aug 15, 2021
edited by asinus  Aug 15, 2021